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Transcript of Is 875 wind load
EXPLANATORY HANDBOOK
ON
INDIAN STANDARD CODE OF PRACTICE FORDESIGN LOADS (OTHER THAN EARTHQUAKE)
FOR BUILDINGS AND STRUCTURES
PART 3 WIND LOADS[IS 875 (PART 3): 1987]
I
I
L
BUREAU OF IN DIAN STANDARDSMANAK BHAVAN, 9 BAHADUR SHAH ZAFAR MARG
NEW DELHI 110002
,,”,.
0 BIS 2001
/
SP 64 (S & T) :2001
FIRST PUBLISHED DECEMBER 2001
Q BUREAU OF INDIAN STANDARDS
ISBN 81-7061 -053-4
PRICE : W 650.00
,’, ”
Typeset by Paragon Enterprises, New Delhi 110002
Printed in India
at Viba Press Pvt. Ltd., 122, DSIDC Shed, Okhla Industrial Area Phase-I, New Delhi- I 1002O
Published by
Bureau of Indian Standards, New Delhi 110002
Composition of the Special Committee for Implementation of Science andTechnology Projects (SCIP)
ChairmanPADAMSHRI DR H. C. VISVESVARAYA
Vice-chancellorUniversity of Roorkee
Roorkee
Members RepresentingDR T. V. S. R. APPA RAO Structural Engineering Research Centre (CSIR),
ChennaiDIRECTOR Central Building Research Institute, RoorkeeSHRI V. RAO AIYAGERI Department of Science & Technology,
New DelhiADDITIONAL DIRECTOR GENERAL (S&P) Central Public Works Department, New Delhi
CHIEF ENGINEER (DESIGNS) (Alternate)SHRI S. K. DATTA Metallurgical and Engineering Consultants
(India) Ltd, RanchiSHRI l?. D. MAYEE Planning Commission, New Delhi
SHRI UMESH KALRA (Altenzate)
Member-Secretaries
SHRIMATI NEETA SHARMA
Deputy Director (S&T), BIS
SHRIMATI RACHNA SEHGAL
Deputy Director (Civ Engg), BIS
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As in the Original Standard, this Page is Intentionally Left Blank
FOREWORD
Users of various civil engineering codes have been feeling the need for explanatory handbooks and othercompilations based on Indian Standards. The need has been further emphasized in view of the publication ofthe National Building Code of India in 1970 (which has since been revised in 1983) and its implementation.The Expert Group set up in 1972 by the Department of Science and Technology, Government of India carriedout in-depth studies in various areas of civil engineering and construction practices. During the preparation ofthe Fifth Five-Year Plan in 1975, the Group was assigned the task of producing a Science and Technology Planfor research, development and extension work in the sectors of housing and construction technology. One ofthe items of this plan was the formulation of design handbooks, explanatory handbooks and design aids basedon the National Building Code and various Indian Standards and other activities in the promotion of the NationalBuilding Code. The Expert Group gave high priority to this item and on the recommendation of the Departmentof Science and Technology, the Planning Commission approved the following two projects which were assignedto the Bureau of Indian Standards (erstwhile Indian Standards Institution) :
a) Development programme on code implementation for building and civil engineering construction, and
b) Typification for industrial buildings.
A Special Committee for Implementation of Science and Technology Projects (SCIP) consisting of expertsconnected with different aspects was set up in 1974 to advise the BIS Directorate General in identifyingthe handbooks and for guiding the development of the work. Under the first project, the Committee hasidentified several subjects for preparing explanatory handbooks/compilations covering appropriate IndianStandards/Codes/Specifications which include the following :
*Handbooks Published
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Il.
12.
13.
Design Aids for Reinforced Concrete to IS 456:1976 (SP 16: 1980)
Handbook on Masonry Design and Construction (first revision) (SP 20: 1991)
Summaries of Indian Standards for Building Materials (SP 21: 1983)
Explanatory Handbook on Codes of Earthquake Engineering (IS 1893:1975 and IS 4326: 1976)(SP 22: 1982)
Handbook on Concrete Mixes (SP 23: 1982)
Explanatory Handbook on Indian Standard Code of Practice for Plain and Reinforced Concrete (IS 456:1978) (SP 24: 1983)
Handbook on Causes and Prevention of Cracks in Buildings (SP 25: 1984)
Handbook on Functional Requirements of Industrial Buildings (Lighting and Ventilation) (SP 32:1986)
Handbook on Timber Engineering (SP 33: 1986)
Hankbook on Concrete Reinforcement and Detailing (SP 34: 1987)
Handbook on Water Supply and Drainage with Special Emphasis on Plumbing (SP 35: 1987)
Handbook on Typified Designs for Structures with Steel Roof Trusses (with and without Cranes) (basedon IS Codes) (SP 38: 1987)
Handbook on Structures with Steel Portal Frames (without Cranes) (SP 40: 1987)
,
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* Hmdbookspublishedare available for sale from BIs Headquarters, and from all Branches and Regional, Offices of BIS.
(v)
.,
14. Handbook on Functional Requirements of Buildings (Other than Industrial Buildings) (SP 41: 1987)
15. Handbook on structures with Reinforced Concrete Portal Frames (without Cranes) (SP 43: 1987)
16. Handbook on Structures with Steel Lattice Portal Frames (without Cranes) (SP 47: 1987)
17. Handbook on Building Construction Practices (Other than Electrical Services) (SP 62: 1997)
18. Handbook on Construction Safety Practices (SP 70: 2001)
This Handbook has been written with a view to provide detailed background information on the provision ofIndian Standard on Code of Practice for Design Loads (Other than Earthquake) for Buildings and Structures :Part 3 Wind Loads [1S875 (Part 3): 1987] and also the use of standard for arriving at the wind loads on buildingsand structures while evaluating their structural safety. The Handbook will serve as a guide for all those engagedin the structural design of wind sensitive buildlngs and structures due to their shape, slenderness, flexibility, sizeand lightness.
It maybe noted that the Handbook does not form part of any Indian Standard on the subject and does not havethe status of an Indian Standard. Wherever, if there is any dispute about the interpretation or opinion expressedin this Handbook, the provisions of the codes only shall apply; the provisions of this Hankbook should beconsidered as only supplementary and informative.
The Handbook is based on the first draft prepared by Shri S.K. Agarwal, Head, Wind Engineering, SERC,Ghaziabad. The draft Handbook was circulated for review to STUP Consultants Ltd, Mumbai; National Councilfor Cement and Building Materials, New Delhi; Central Public Works Department, New Delhi; Indian Instituteof Technology, Chennai; Indian Institute of Technology, KanpuG Structural Engineering Research Centre,Chennai; M.N. Dastur & Co Ltd, Kolkata; Meteorological Office, Pune; Indian Institute of Science, Bangalore;University of Roorkee, Roorkee; Tandon Consultants Pvt Ltd, New Delhi; Central Electricity Authority, NewDelhi; Indian Institute of Technolo~y, Kharagpuq Indian Institute of Technology, New Delhi; Engineers IndiaLtd, New Delhi; N.T.P.C., Noida; Ministry of Railways (RDSO), Lucknow; Department of Space, Bangalore;Nuclear Power Corporation, Mumbai; Moti Lal Nehru Regional Engineering College, Allahabad; VaJcilMehtaSeth, Ahmedabad; Anna University, Chennai; UNITECH, New Delhi; Consulting Engineering Services (I) PvtLtd, New Delhi; Tata Consulting Engineers, Mumbai; Housing &Urban Development Corporation, New Delhi;National Institute of Construction Management and Research, Mumbai; Howe India (Pvt) Ltd, New Delhi; ArmyHeadquarters, New Delhi; Institution of Engineers, New Delhi; Gammons India Ltd, Mumbai; Central BuildingResearch Institute, Roorkee; Department of Science & Technology, New Delhi; Metallurgical& EngineeringConsultants (India) Ltd, Ranchi; Planning Commission, New Delhi; and views expressed by them were takeninto consideration while finalizing the Handbook.
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(vi)
ContentsClause
O INTRODUCTION
0.1 Background
0.2 Nature of Wind
0.3 Shortfalls of the1964 Code
1 SCOPE
2 TERMINOLOGY
2.1 Angle of Attack
2.2 Breadth
2.3 Depth
2.4 Developed Height
2.5 Effective Frontal Area
2.6 Element Surface Area
2.7 Force and Moment Coefficients
2.8 Ground Roughnes
2.9 Gradient Height
2.10 Pressure Coefficient
2.11 Suction
2.12 Solidity Ratio’@’
2.13 Fetch Length
2.14 Terrain Category
2.15 Velocity Profile
2.16 Topography
3 WINDSPEEDANDPRESSURE
3.1 Basic Wind Speed
3.2 Design Wind Speed3.3 Risk Coefficient
3.4 Terrain, Height and Structure Size (k2Factor)
3.5 Fetch and Developed Height Relationship
3.6 Topography (ks)
3.7 Design Wind Pressure
4 WINDPRESSUREANDFORCESONBUILDINGS/STRUCTURES
4.1 General
4.2 Pressure Coefficients
4.3 Force Coefficients
5 DYNAMICEFFECTS
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6 GUSTFACTOR(Go ORGUSTEFFECTIVENESSFACTOR(GEF,)METHOD..
7 WINDTUNNELMODELSTUDIES ...
8 APPLICATIONOFCODALPROVISIONS ...
9 ASSESSMENTOFWINDLOADSONBUILDINGSANDSTRUCTURES ...
LIST OF SOLVED EXAMPLES ON
Rectangular Clad Buildings ...
Free Roof ...
Miscellaneous Structures (Force Coefficient Method) ...
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O INTRODUCTION
0.1 Background
A large number of structures that are being constructedat present tend to be wind-sensitive because of theirshapes, slenderness, flexibility, size and lightness.Added to these are the use of a variety of materialswhich are stressed to much higher percentage of theirultimate strength than in earlier days because of betterassurance of the quality of materials. In the socialenvironment that is developing world over, the ancientphilosophy of accepting continuing disasters due towind as ordained by ‘fate’ and Gods is giving place todemands for economical wind resistant designs. Thesefactors have demanded a more realistic, if not, precisedefinition of wind loading. Updating of someInternational Codes of practice, notably the British,Australian, Canadian, American and French has beeneffected fairly frequently over the last two decades andthe present versions incorporate most of the advancesmade in understanding the wind characteristics and itseffect on structures. The new discoveries are such thatit is clear that mere issue of amendments to the earlierwind Code IS 875 : 1964 will not be justifiable. Therecently issued wind code ‘Code of practice for designloads (other than earthquake) for buildings andstructures’ IS 875 (Part 3): 1987 differs in many waysfrom the previous Code first issued in 1964 andattempts not only to rectify the shortfalls of the 1964code but incorporates recent knowledge of windeffects on structures. The height up to which velocitiesare given has now been raised to 500 m and theloadings on as many of the commonly encounteredbuildings and structures, for which there are no otherIndian Standards, have been included. Although notexplicitly stated, the code recognizes the fact that mostof the high winds in India occur due to short durationrotating winds like tropical cyclones along the Coastsor Tornadoes elsewhere, and nearly rectilinear windsof short duration like thunderstorms at many places. Inthis respect, the high wind loading conditions in Indiaare different from those of temperate zone countrieslike Europe and Canada. Much of the random responsetheories, which have been adopted in EuropanA-J.S. orAustralian Codes are based on these ‘fully developedpressure winds’ or ‘pressure wind’ conditions andstrictly cannot be applied in most parts of India. Buttheir judicious use, in the absence of proper theoriesapplicable to cyclones, tornadoes and thunderstormswill give adequate safety margins and this is what thepresent 1S 875 (Part 3) :1987 attempts to do.
In India, success in satisfactory codification of windloading on structures has remained elusive so far. Inmost cases, codification has followed, not precededstructural failures or distress. The wind maps given inthe 1964 version of the code had been prepared mainl yon the basis of extreme value wind data from storms
SP 64 (S & T) :2001
which approached or crossed the Indian coasts during1890 to 1960, together with the wind data availablefrom about 10-12 continuously recording Dynes Pres-sure Anemograph (DPA) stations which existed at thattime, to get an overall picture for the country. However3-cup anemometer readings were not much used in thepreparation of wind maps because much of such datawere synoptic.
The height of DPA instruments varied from 10 m to30 m at different places. Therefore only one extremevalue of wind was given up to 30 m height from groundlevel without any variation in-between. Further, l/10thpower law had been adopted regardless of terrainconditions, for indicating variation of wind speed withheight from 30 m to 150 m, for which there was nosupportive evidence. The code gave two wind pressuremaps (one giving winds of shorter duration <5 minutesand another excludlng winds of shorter duration) andthere was no clear guidance for using either or both ofthem.
With the publication of the recent revised wind code,IS 875 (Part 3) : 1987, an attempt has been made toremove these deficiencies and provide to the Indianstructural engineer adequate guidelines for arriving atmore rational wind loading for design purposes.
0.2 Nature of Wind
0.2.1 Wind means the motion of air in the atmosphere.The response of structures to wind depends on thecharacteristics of the wind. From the point of view ofassessing wind load, it is convenient to divide the windinto two categories: ‘Rotating and Non-rotating’winds. Rotating winds are caused by tropical cyclonesand tornadoes. The wind speeds caused by these mayexceed 200 kti. The duration of such winds at anyIoeation varies from 2 to 5 minutes in the case oftornadoes and thunderstorms and about 3 to 4 days incase of tropical cyclones. Non-rotating winds arecaused by differential pressures and thus move inpreferred direction. These are also called ‘PressureSystem’ winds and when they persist for distances like50-100 km, are termed as ‘Fully developed pressuresystem winds’. The intensity of such winds is usual]ygiven in Beaufort Scale as in Table 1 (other scales forparticular types of winds like cyclones have also beenproposed). Thunderstorms are rectilinear winds ofhigh speed lasting only a few minutes and start as astrong vertical downdraft, from clouds which spreadlaterally on reaching the ground. They last for 2 to5 minutes and wind speed can exceed 200 km/h.
0.2.2 Gradient Winak
Winds are primarily caused by the differences intemperature over the earth’s surface because theintensity of the sun’s radiation received at different
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SP 64 (S & T) :2001
latitudes varies, and land areas heat and cool quickerthan the sea. Ocean currents also affect the temperaturedistribution by transporting heat from one part of earthto another. The differences in temperature give rise tothe gradients of pressure which set the air in motion.Near the earth’s surface, the motion is opposed, andthe wind speed reduced, by the surface friction. At thesurface, the wind speed reduces to zero and then beginsto increase with height, and at some height, known asthe gradient height, the motion may be considered tobe free of the earth’s frictional influence and will attainits ‘gradient velocity’. This velocity depends on thepressure variation at that height in the plane parallel tothe earth’s surface. The gradient height increases withthe roughness of the terrain on the earth’s surface andis taken to vary from about 300 m for flat ground toabout 550 m for very rough terrain, in the Indian anda few other national codes. Although, the gradientheight is confused in many codes to be the same as theedge of the earth’s (planetary) boundary layer, the twoheights are not identical. True gradient height lies wellabove the edge of the earth’s boundary layer height.The Indian Code has adopted a carefully writtendefinition to imply this difference.
0.2.3 Variation of Mean Wind Speed with Height
At the earth’s surface the wind speed is zero and, inconditions of neutral stability, there is a continuousincrease of mean wind speed from the ground to thegradient height. A number of laws for defining thiswind speed profile have been suggested, the mostwidely used of which is the logarithmic profile. Theempirical power law profile is also used in many codes.
0.2.3.1 Logarithmic profile
The logarithmic profile of the variation of mean windspeed with height is given by
V~F = Ilk. lo& zI@ ...(1)
design wind velocity at height z,von Karman’s constant with a numericalvalue of 0.4,height above groundis the surface roughness parameter,the friction velocity defined as ~,the skin friction force on the wall, and
the density of air.
0.2.3.2 Power luw
Another representation for the mean wind speedprofile is the simple power law expression:
VziVR= (?,kR)a ...(2)
whereVR = the speed at a reference height ZR, usually
taken as the standard meteorological heightof 10 m, and
a = a constant for a particular site and terrain.
0.2.4 Nature of Wind Flow Past Bodies
0.2.4.1 The nature of flow of a fluid past a bodyresting on a surface depends on the conditions of thesurface, the shape of the body, its height, velocity offluid flow and many other factors. The flow may be
Table 1 Beaufort Scale for Wmd Speeda(Clause 0.2.1)
Beaufort Description speed Deacrtptionof Wind EffectsNo. of Wind dso Calm < r3.4 No noticeablewind1 Lightairs 0.4-1.5 Barelynoticeablewind2 Lightbreeze 1.6-3.3 Whrdfelton face3 Gentlebreeze 3.4-5.4 Windextendstightflag,hair is distuti, clothingflaps4 Moderatebreeze 5.5-7.9 Windraisesdust,dry soil,loosepaper,hairdisarranged
5 Freshbreeze 8.0-10.7 Forceof windfelt on body,driftingsnowbecomesair borne;timitof agreeablewindon land
6 Srrongbreeze 10.8-13.8 Umbrellasusedwithdifficulty;hair blownstraigh~dlftlcultyin watking steadily;windnoiseon earsunpleaaan~begin-ningof blizzards (snowflowsin air)
7 Moderategale 13.9-17.1 Inconveniencefeltwhenwatking
8 Freshgale 17.2-20.7 progressinwalkingdiffIculcmaintainingbalanceingustsverydifficult
9 Stronggale 20.8-24.4 Peopleblownoverby gusts10 Wholegale 24.5-28.5 smalltreesuprooted11 Storm 28.6-34.0 Widespread&mage
12 Hurricane >34.0 Failureof illdesigned, structures,likelampposts,GI sheetstomajorfailures
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what is called ‘streamlined’ or ‘laminar’ flow or whatis called ‘turbulent’ or ‘unsteady flow of randomnature’. A streamline is defined as a line drawn througha moving fluid such that the velocity vector istangential to it. The streamlines coincide with the pathsof the particles in steady laminar fluid motion but notwhen the fluid motion is turbulent. In turbulent fluidmotion, the velocity and direction of the particles ofsmall masses of fluid fluctuate in time, mostly in arandom and chaotic manner. In uniform flow eachstreamline has the same constant velocity, but in flowsof viscous fluids near a solid boundary, the velocity atthe boundary is reduced to zero by the effects ofviscosity. Thus the velocity along the streamlinesincreases with distance of the streamline from the solidboundary until a final unretarded velocity is attained.The region of retarded fluid, which is indicated by thebroken lines in Fig. 1 is referred to as the boundarylayer. The flow over the surface within the boundarylayer may be laminar at an upstream position buttransition to turbulent flow may take place at someposition as the flow proceeds downstream along thesurface. This transition takes place over a distancealong the flow direction. If the flow is subject to anadverse gradient, that is a pressure gradient opposingit (such as may be induced by flow over a concavesurface, or other circumstance in which it is forced orallowed to expand), its velocity throughout theboundary layer is further retarded and at some positionalong the surface may be reduced to zero slightlyabove the surface. Then the opposing pressure willdrive the fluid in opposite direction (reversed flow).The boundary layer flow becomes detached from thesurface and large eddies are formed and are dischargedinto the region behind the body called the wake. Theseeddies are usually discharged either randomly or atfairly regular intervals, dependtng on the vebcity ofthe fluid and the dimensions and shape of the body.The energy carried downstream by these eddiesreduces the pressure in the wake and hence the
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SP 64 (S & T) :2001
pressures acting on the body surfaces in contact with‘it, and this reduction of pressure on the body’s leewardside accounts for a large proportion of the drag force.Pressure forces act normal to the surface, andintegrating the components of pressure forces over thewhole surface of the body in-line and perpendicular tothe fluid direction results in alongwind (or drag) andacrosswind (or lift) forces. These are additional to theskin-friction drag, which is caused by the viscosity ofthe fluid and acts tangentially to the surface. The dragresulting from the pressure distribution is referred toas the pressure drag, and is kept to a minimum inshapes used in aircraft wings called aerofoil sections,which are designed to avoid flow separation.
Except for surfaces with pronounced protuberancesthat may fix the positions where the flow separates, thepositions of transition from laminar to turbulent flowand of flow separation from the surface are dependenton the Reynolds number (Re) of the flow,
R== pvl~ = WV ...(3)
where
p and v are, respectively, the dynamic and kinematicviscosities of the fluid and 1is a typical length.
The Reynolds number is a measure of the ratio of theinertia forces (that is, force due to acceleration of fluidparticles) to viscous forces in a fluid.
0.2.5 Flow Patterns
0.2.5.1 Around circular cylinders
We will first consider flow pasta structure of circularcross-section, since this shape is used in many tall andeven short structures. The flow patterns around acircular cylinder are sketched in Fig. 2, and show themarked influence of Reynolds number (Q. In thesubcritical flow regime Fig. 2(a) the boundary layer
FIG. 1 GROWTHOFABOUNDARYLAYERINA FAVORABLEPRESSUREGRADIENT
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5P 64 (S & T) :2001
remains laminar up to the points of separation, whichoccur at about 80° from the stagnation point P.Vortices are formed regularly and alternately fromeach side of the cylinder and are shed into the wake.The width of the wake, and hence the value of the dragcoefficient Cd, is greatest in the subcritical flowregime. When the surface is smooth, at about aR, = 2 x 105, the laminar boundary layer becomesturbulent at a forward position on the cylinder surface.The separation point shifts more or less suddenly to anangle of about 120° to 130°. In this critical regimeFig. 2(b) cd falls rapidly as the Reynolds numberincreases to 5 x 105when its lowest value is reached.Thereafter, increase of Re brings the flow into thesupercritical regime Fig. 2(c) with an increase in the
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LB =T=TA =LS, TS =TB =
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width of the wake and an increase in cd. The periodicvortex shedding in this regime is both weak andrandom. The increase in cd continues until Re = 3 X 106and the position of the separation point stabilizes atabout 109°. In :his transcritical (sometimes, perhapsmore correctly, referred to as hypercritical) regimeFig. 2(d) the value of cd decreases slowly with increaseof Reand the wake contains peak energy at frequencieshigher than in the subcritical regime vortex sheddingfrequency. The flow patterns of Fig. 2 are forstationary smooth-surface cylinders in a smoothapproach flow; they are modified by surfaceroughness, by turbulence of the incident wind, and bymotion of the cylinder and height to diameter ratio ofthe cylinder.
-a) -> J -m
c’ —————___fbl
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Laminar Boundaty LayerTransition from Laminar to Turbulent Boundaty LayerRe-attachment of Turbulent Boundary LayerSeparation of Laminar and Turbulent Boundaries RespectivelyTurbulent Boundaty Layer
Broken line indicates boundary of separated flow.
(a) Subcritical, 300 c R, <2 x 105, Q= 1.2 a= 0.2; (b) Critical,2 x 105 c R. c 5 x 105,1.2> Cj >0.3,0.2< S, c 0.5; (c) Super-critical, 5 x 10b < Re c 3 x 106, 0.5s ~s 0.7, S = 0.5; (d) Transcritical(Hypercritical), R, >3 x 1C6,Q = 0.7, S = 0.27, R,= V&.1,where dis the diameter of the cylinder
FIG. 2 SKETCHESOFTWO-DIMENSIONALAIRFLOWPATTERNSAROUNDA CIRCULARCYLINDER
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0.2.5.2 Around other bluff bodies
A body is said to be a bluff body if the ratio of itsdimension parallel to the flow to its dimensionperpendicular to the flow does not differ by more thanabout 6 and whose drag coefficient based on thesmaller dimension exceeds about 0.1. Circularcylinder, square and rectangular prisms, etc, areexamples. As an example, the flow patterns around along square-section prism are sketched in Fig. 3. Thepatterns, and hence the values of the drag coefficientCdcan be considered independent of R, if R, is > about104because the positions of the points of separation ofthe flow from the body are fixed by the sharp corners.They are, however, influenced by the turbulence of theapproaching stream, the effect of which may be tocause the flow which has become detached at awindward comer (A) to reattach to the side, untilfinally separated at the leeward comer. Thus inturbulent flows the width of the wake is reducedcompared to that in smooth flow, with a correspondingreduction in the value of Cd.
<u -,.-{-.———----—-.... .
.-77-’A. --_______f---.-.c
—-—__Ibl
(a) Smooth Incident Flow, Cd= 2.2; (b) TurbulentIncident Flow, Cd= 1.6
FIG.3 SKETCHESOFAIRFLOWPATTERNSAROUNDPRISMSOFSQUARESECTIONWITHSHARPCORNERS
0.2.5.3 Over buildings
Figure 4 shows sketches of typical flow patternsaround buildings, most of which can be classed as bluffbodies.
0.2.6 Wind Effects on Structures
Wind effects on structures can be classified as ‘Static’and ‘Dynamic’. Strictly speaking, wind effects areonly dynamic in nature but in most of the codes thesedynamic effects are expressed in terms of an
SP 64 (S & T) :2001
equivalent static load. Static wind effect primarilycauses elastic bending and twisting of structure. Thedynamic effects of wind are either periodic forces suchas due to Vortex Shedding, Flutter, Galloping andOvalling or non-periodic such as turbulent bufetting(see Table 2).
0.3 Shortfalls of the 1%4 Code
0.3.1 The Code, which was published more than 35years ago has provided reasonably good guidance tostructural engineers for the design of most of thesimple structures. In most cases, its provisions havebeen safe, if not conservative.
A look at some of the recent International Codeshowever, clearly indicates the significant advancesmade in wind engineering in the last few years. Notonly is there an improved knowledge of the charac-teristics of wind, but there have been new trends,innovations and requirements of structural design,which demand a more accurate definition of windforces. Viewed from this background, the shortfalls ofIS 875:1964 must be understood in today’s context.In the revision of the Code, an attempt has been madeto overcome the obvious shortfalls and update theinformation to present-day knowledge of windengineering. A serious attempt has been made toensure that the new phenomenon introduced or/anddiscussed in the 1987 revision are free from am-biguities and not significantly beyond the existingknowledge base in the country. Some of the moreimportant shortfalls that had surfaced over the years ofthe Code of 1964, which provided the main impetus tothe revision of the 1964 Code are briefly summarizedbelow:
0.3.2 Wind Zoning
Of interest to structural designers are wind pressuresof short averaging time and gusts with averaging timeof say 3 to 15 seconds. The wind map zoning in the1964 Code represented the state of knowledge of thewind climate and measurements taken earlier oversome 30 years. Several studies have since beenpublished based on increasingly available data. It wasabundantly clear that the earlier wind zoning eitherover-estimated or under-estimated the occurrence ofextreme wind speeds in different parts of the country.Thus the revision of wind map zoning was longoverdue.
0.3.3 Risk Level
In the 1964 Code, no indication was available aboutthe return period. Experiences all over the world, byand large, has favoured designing normal structuresfor a life and return period of 50 years as these criteria,seem to yield safe and economical loads for normalstructures. For wind-sensitive structures and for
5
SP64(S&T):2001
(0)
—*-—
(b)
(d)
(a) Flat roof H/b = 0.5, Flow Re-attaches; (b) Flat Roof Mb = 2, No Flow Re-attachment; (c) Low-pitch Roof,Flow Re-attaches; (d) High Pitch roof, No Flow Re-attachment
FIG. 4 SKETCHESOF FLOW PAST LOW-RISE BUILDINGSSHOWINGTHEEFFE~S OFALONG-WINDWIDTHANDOFROOF PITCHONTHE SEPARATEDREGIONINTHEWAKEOFTHEBUILDING
Table 2 Wind Effects on Structures(Clause 0.2.6)
I WindEffects I
1 II I1 I
1
DuetoAtmosphericTurbulence
!
DuetoBuffeting by
TurbulentWalaof other
structures
Ivortex
Excitation(Slender
Structures)likeChimneys,
CoolingTowers,Stacks,BridgeDecks,
etc
I
Buffeting Dueto Periodic
Wakeof otherstructures T
Ovalling Gatloping(TldnWalled (Non-circular
Structures)like Chimneys,Chimneys, Conductors,
Cooling CableBridge,Towers,etc Towers,etc)
IFlutter
(BridgeDecks,AntennaDkh)
f
,,
6
structures with post-disaster functions, the returnperiod is often increased to 100 years. As a matter offact, for structures of exceptional importance thereturn period could logically be even higher to reducethe risk levels still further. These issues are furtherdiscussed in 3.3.
0.3.4 Wind Velocity Measurement
The wind instruments used by meteorologists aregenerally calibrated in terms of wind velocity atstandard atmospheric conditions of 15 ‘C and 760 mmmercury pressure and measure the wind velocity withan averaging time of about 3 seconds . The windpressure is evaluated by the equation p = K@, whereK is a constant. The Code of 1964 gives wind pressureand these were arrived at from the velocities by usinga value of K which is higher by 25 percent as comparedto other International Codes.
0.3.5 Wind Variation with Height and Terrain
The variation of wind velocity with height andcharacter of surrounding terrain has importantimplications in structural design. The 1964 Codespecified constant pressure up to a height of 30 m andthereafter, a variation corresponding to the followingequation was given:
Pz = P30 (zj30)a ...(4)
wherePyj =
z=
a=
basic wind pressure which is constant up toheight of 30 m;height in m above ground of point underconsideration; andpower index, with a value of 0.2.
Whenever such power laws are used, the generallyaccepted norm in International Codes is to take aconstant velocity up to 10 m height and thereafter avariation given by the following equation to obtain thevelocity profile:
Vz = v~~(Z/lo)a ...(5)
whereV,() =z =
a=
velocity at 10 m above ground;height in m of point above ground underconsideration; andpower law index whose value varies from0.07 to 0.35 depending upon the terrain andaveraging time; larger indices correspond toincreasing roughness of terrain and largeraveraging times.
In recent times, the logarithmic law is being used moreoften, mainly due to its better theoretical backing.Considering that the power law index of IS 875:1964was applicable to pressures, the corresponding indexfor velocity would be 0.1. While structures of low
7
SP 64 (S & T) :2001
height may not be affected significantly by the lowvalue of the power index, the 1964 Code would yieldpressures which are under-estimates in the upperregions of tall structures where more precise values areneeded. The Code did not indicate variations withterrain roughness, and was silent about wind pressuresabove 150 m from ground. With tall structures alreadyreaching heights of 300 m in the country and still tallerstructures on the drawing board, the short falls ofIS 875 : 1964 became particularly evident for tallstructures.
0.3.6 Dynamic Analysis
For tall, long span and slender structures a ‘dynamicanalysis’ of the structure is essential. Wind gustscause fluctuating forces on the structure which inducelarge dynamic motions, including oscillations. Theseverity of the dynamic motions and/or wind-inducedoscillations depends on the natural frequency ofvibration and the damping of the structure. Themotions are induced in both ‘along-wind’ direction aswell as ‘across-wind’ direction. The ‘along-wind’response of the structure is accounted for by amagnification factor (often called the ‘gust factor’ )applied to the static or direct forces. The ‘across-wind’response requires a separate dynamic analysis. The1964 Code did not give any guidance on these aspectsof structural design of tall structures.
0.3.7 Size of Structure
The 1964 Code did not indicate any correlationbetween the structure size and wind loading. Themagnitude of the load imparted on a structure by a gustdepends, to a significant extent, on the size of thestructure. The reason is that the spatial extent of a gustof wind in any directions, having the same velocity asthat specified in the Code is small-being about 25 m to150 m at any instant of time. Thus, a large structurewill not experience the highest wind specified at allpoints on it at the same instant and hence the averagetotal load on the structure will be less than the valueobtained by assuming the highest specified wind overits entire extent. Allowance for this needs to be made,although locally the load will be that due to the highestwind.
0.3.8 Shape of Structure
Structures can be of many shapes, particularly whenwe combine the whole gamut of possibilities inconcrete and structural steel. The ‘direct or static’ windforce on a structure is given by the equation F = SAP,where S is the shape facto~ A is the projected area inthe plane perpendicular to the wind direction; and P isthe wind pressure. The 1964 Code gave values of theshape factor for the basic shapes (circular, octagon andsquare), but this information needed to be extended toother shapes for various wind directions. For steelstructures of different plan shapes, shielding
SP 64 (S & T) :2001
configurations and solidlty ratios, no guidance wasavai Iable. Therefore it became essential for thedesigner to either consult other International Codes ortake advice from specialists.
0.3.9 Pressure Coefficients
All over the world, the unsatisfactory performance ofrelatively light weight low rise buildings such as thethatched houses in rural India or houses with sheetroofing, led to updating of pressure coefficients(internal and external) for various configurations ofbuildings, wall openings and wind directions.Additionally, a need was felt to define higher localexternal negative coefficients at certain locations tocater to concentrations near edges of walls and roofs,where the failures are primarily initiated. It isrecognized that considerable research is still needed inthis area.
1 SCOPE
The aim of this Handbook is to provide detailedbackground information on the provisions of IS 875(Part 3) : 1987 and the use of the standard forestimating wind loads on structures. The emphasis hasbeen on the identification of source material used,philosophical ideas behind the changes in 1987 editionand guidance to the use of the standard. Clauses 0.1and O.2 of this Handbook contains backgroundinformation, philosophical ideas, nature of wind andwind effects on structures. Clause 0.3 brings out theshortfalls of the earlier standard (IS 875: 1964),vis-a-vis wind zoning, risk level, wind velocity and itsvariation with height, dynamic analysis, effect ofstructure size and shape. Clause 2 explains the variousterminology used in the standard. Clause 3 discussesthe main features of the revised Code like new windzoning map, estimation of design wind speed, conceptof return period, effects of terrain roughness andstructure size, local topography, correctvelocity-pressure relationship. In Clause 4 windpressure and forces on buildings/structures arediscussed in light of pressure coefficient method.Details of force coefficient method are given in 4.3while dynamic effects like along-wind andacross-wind responses are explained in 5 and 6.Requirements of wind tunnel studies are detailed in 7.Clause 8 conducts the user to the various provisions ofthe standard with the help of flow diagrams. Clause 9contains the methodology for the estimation of windloads on low rise buildings, with the help of examples.Clause 9.3 is devoted to the estimation of dynamiceffects of wind using gust factor method and containstypical solved examples of structures.
NOTE—AllClauses,FiguresandTablesinSquareBracketsinthe HandbookRefer to ClauseNumbers,FigureNumbersandTableNumbersof IS 875(Part3): 1987,respectively
2 TERMINOLOGY
2.1 Angle of Attack [3.1.1]
Angle between the direction of wind and the referenceaxis is known as the angle of attack. As shown inFig. 5(a), angle %’ is angle of attack, between thedirection of wind and reference axis (horizontal line).
2.2 Breadth [3.1.2]
Horizontal dimension of the building measured normalto the direction of wind, is termed as the breadth of thestructure. As shown in Fig. 5(b), breadth =AB for windalong the longitudinal reference axis and BK for windperpendicular to it. For other wind directions, such asat angle ’0’, it is the maximum width seen along thewind such as the projection of AB’ on a planeperpendicular to the direction of wind. In Fig. 5(b),this is AE.
2.3 Depth [3.1.3]
Horizontal dimension of the building measured in thedirection of wind is the depth of the building. As shownin Fig. 5(b), depth = AA’for wind along the referenceaxis and BA for wind perpendicular to the referenceaxis. For other wind directions, such as at angle ‘f3’,itis the maximum length from the first point on the bodywhich meets the wind to the point where a line fromthis point along-wind meets a boundary of the bodysuch as BD.
2.4 Developed Height [3.1.4]
The velocity increases up to certain height from theground when wind passes from one terrain categoryto another terrain category. After covering a certaindistance in the new terrain, the velocity profile doesnot change any more and is said to be fully developed.Before becoming fully developed, the velocity profilewill have a mixed character of the fully developedprofiles of the upstream and downstream terrains. Inthis intermediate region, the height up to which thevelocity profile changes in the new terrain is known asthe Developed Height. [Table 3] gives the vahtes of theDeveloped Height as a function of the distance for thefour terrain categories. Methodology for estimation ofthe developed height and for using the same has beenillustrated in the solved Examples given in 3.5.
2.5 Effective Frontal Area [3.1.5]
As given in Fig. 5(a) for angle of attack ’6’ = O,effective frontal area is ABCD, which is the projectedarea of the structure normal to the direction of thewind. For ‘til’not equal to zero, the frontal area isACEF.
2.6 Element Surface Area [3.1.6]
In Fig. 5(a), for angle of attack ‘f3’= Oat height z, areaPQRS represents the element surface area of heightAz. In the element surface area pressure/forcecoefficients are to be taken constant.
,,
8
(
I
!.,4s. 9 &
2.7 Force and Moment Coefficients [3.1.7]
When a structure is immersed in flowing fluid like air,a force is exerted on it which depends on the nature offlow, type of structure and is expressed in terms ofnon-dimensional coefficient. This non-dimensionalcoefficient is termed as force-coefficient.
There can be components of force in the directionsother than the direction of wind. The force componentalong the direction of wind is called drag force (now-a-days, more frequently as along-wind force) and theforce components in the other normal directions aretermed as lift force (or across-wind force) and sideforce (or transverse force). Likewise, the momentcoefficient tending to bend the structure in the direc-tion of wind is called the over turning moment, thecoefficient tending to twist it about a vertical axis astorque and the coefficient in the transverse direction as‘sideways’ moment coefficient.
2.8 Ground Roughness [3.1.8]
The obstructions like trees, buildings, structures, etc,on ground are termed as ground roughness. The groundroughness tends to retard the flow near ground becauseof momentum transfer between layers of wind. Thevelocity profile of wind is dependent on the groundroughness. Although the nature of ground roughnesson the surface of the earth varies from flat and verysmooth to very rough as in the city centres with tallbuildings, it has been found convenient and adequateto categorize them into four fairly distinct categories.
/’7
SP 64 (S & T) :2001
2.9 Gradient Height [3.1.9]
When wind moves over large distances in a particularterrain, the velocity becomes constant above a certainheight. The height above which the velocity is foundto be constant and is not influenced by surface frictionis called the Gradient Height. Implicitly, the Codeidentifies the gradient height as the height of theplanetary boundary Iayev the wording of the definitionis obviously very carefully done, since themeteorologists’ identify the gradient height as one atwhich there is a balance of Coriolis and pressureforces. The meteorologists’ definition of gradientheight puts it well above the edge of the atmosphericboundary layer. Thus the definition of gradient heightused in the Code is not the same as the gradient heightas understood by the meteorologists.
2.10 Pressure Coefficient [3.1.9]
The pressure exerted by wind at a point on the structurewil I be different from the pressure of wind farupstream of the structure, called ‘free stream staticpressure’. The ratio of the difference of the pressureat a point on the structure and static p~ssum of theincident wind to the design wind pressure is termed asPressure Coefficient.
The concept of pressure coefficient is similar to that offorce coefficient. The pressure coefficient at a pointmultiplied by design wind pressure yields the pressureat that point.
J
ee“RENCE
AXIS
WIND DIRECWM
#
(e) (b)VJINDIN PLAN BUILDINGPLAN
FIG. 5 WIND EFFECTSONA BUILDING
9
SP 64 (S & T) :2001
The pressure coefficient may vary with height orwidth. The pressure acting towards the surface ofstructure is to be taken as positive and pressure actingaway from the surface of structure is to be taken asnegative.
2.11 Suction [3.1.9]
Atsomepoints on anybody, pressure may fall belowatmospheric pressure and is called ‘suction’. Suctionis the pressure acting away from the surface ofstructure; negative pressure is the same as suctionpressure.
2.12 Solidity Ratio ‘$’ [3.1.9]
In general, solidity ratio is the ratio of the sum of allthe areas’ of each element of a framed structurenormal to the wind to the area of the boundaryenclosing the structure. Care must, however, be takenin applying “this definition. In a plane frame, theboundary area is to be taken as that of the main frameonly and the areas of appurtenances which are withinthe boundary [as AntennaAl, or ladderLl, in Fig. 6(a)]treated as part of the element of the frame. However,the area of the antenna or an appurtenance outside theboundary such as the AntennaA2 or lightening arrester
6(a) Plane Frame
0’ c’
(i)(ii)
,,I >-
6(b) Space FrameFIG.6 SOLmITYRATIO— PLANEANDSPACEFRAME
10
L~ in Fig. 6(a) should not be taken into account eitheras part of the enclosed boundary or that of the framein computing the solidity ratio. The loads on suchappurtenances outside the enclosed boundary must beseparately estimated and added to the load on the frameat the appropriate point. In Fig. 6(a), the area enclosedby ABCD k the boundary area and the area of shadedelements to be considered for estimating the solidityratio is indicated. Similar principles apply in the caseof space frames. In case of cross bracings, theprojected area as seen from windward face shall betaken and added to other areas (shown shaded) toarrive at the total solidity ratio. If there areappurtenances within the enclosed volume (ABCDA’B’C’D’) [see Fig. 6(b)(i)], then their area should beprojected on the nearest windward frame forestimating the frame solidity ratio Fig. 6(b)(ii). Thesolidity ratio of each frame should be found andappropriate shielding factor applied.
2.13 Fetch Length [3.1.9]
The distance over which wind has moved in aparticular terrain category before approaching thestructure is known as the Fetch Length. The velocityprofile of wind changes continuously over the fetchlength before stabilizing at the variation given in[Table 2]. The distance required for a stab]: profile tobe formed is given in [Table 3]. Distance AB in Fig. 7is the fetch length.
2.14 Terrain Category [3.1.9]
Buildings, vegetation, walls, trees and waves at seacontribute to the surface roughness and thus influencethe local characteristics to which a structtue may beexposed. The average ground roughness of large areasis termed as Terrain Category. Based on the averageheight of the ground roughness four representativeterrain categories having fully developed velocityprofiles are suggested [5.3.2] from the equivalent of acalm sea to inner city area dominated by tall buildings.
SP 64 (S & T) :2001
Terrain category 1 (TC- 1) applies to exposed openterrain with few or no obstructions and in which theaverage height of objects surroundings the structure isless than 1.5 m. Category 2 (TC-2) refers to openterrain with well scattered obstructions having heightsgenerally between 1.5 to 10m while Category 3 (TC-3)applies to terrain with numerous closely spacedobstructions having size of buildings/structures up to10 m in height with or without a few isolated tallstructures. Terrain Category 4 (TC-4) means terrainwith numerous large high closely spaced obstructions.However, when fetch lengths are small, the velocityprofile is not fully developed and suitable velocityprofile with changes in terrain categories beconsidered [5.3.2.4].
2.15 Velocity profile [3.1.9]
The variation of wind speed with height is called theVelocity Profile. The velocity profile is dependent onthe terrain category, zonal velocity and the fetchlength. Figure 8 shows the velocity profile in the fourterrain categories.
2.16 Topography [3.1.9]
Geographical features such as mountains, hills,escarpments, etc, of an area is known as theTopography of the area in which the structure is builtand this affects the wind speed on and downstream ofthese features.
3 WIND SPEED AND PRESSURE [5]
3.1 Basic Wind Speed [5.2]
In 1960, there existed about 10-12 continuouslyrecording DPA stations in the country compared toabout 50 or so such stations at present. The instrumentscontinuously record wind speed on Y-axis with timeon X-axis and in addition, record the direction of windspeed. Most of these stations are located on the coastalbelt.
H MODIFIEO VELOCITYPROFILE DUE TOCHANGE OFROUGHNESS
Hh:
MEAN GROUNDLEvEL
APPROACHING VELOCITYPROFILE
(a)‘I?==4’(b,
1
i
FIG. 7 EFFECTOFCHANGE INTERRAINCATEGORIES
11
/
. ... .9 e%,,...’
SP 64 (S & T) :2001
The revised wind map is based mainly on the data fromthe anemograph records of 47 DPA stations for periodsfrom 1948 to 1983. All these wind data have beenpublished in India Meteorological Department publi-cations, ‘Indian Weather Review, Annual SummaryPart-A’ published annually from 1948 onwards. Thebasic approach followed for preparing the wind mapis by extracting wind velocities, since analysis of stmc-tures in many cases depends on wind speed directlyand also much of the available data on effects of terrainis in terms of wind speed. The use of wind speeds forstatistical analysis is also in line with the internatio-nallyaccepted practice. The peak values of annualmaximum wind speeds (wind gustiness) for each of the47 stations formed the data for statistical analysis toobtain the extreme wind speeds. In the statisticalanalysis of all the annual peaks at each of the above 47stations in India, IMD used Fisher-Tippet Type-I(Gumbal) distribution. Before analyzing, recordedvalues were reduced to values at 10 m height aboveground for the normalized terrain Category 2. Sincethe data for analyses came from DPA which has anaveraging time of about 3 seconds, analysis gives gustvelocity values averaged over 3 seconds. These valuesfor 50 year return period and 50 year structure life havebeen given in [Fig. 1] for terrain Category 2. In addi-tion to the use of data available from 47 DPA stationsup to 1983, the following effects have been suitablyincorporated in the preparation of the basic windvelocity map:
Orography — Orography has been considered in thezoning of wind velocity map since it was observed thatcontinuous hills and mountains spread over a largearea affect wind speed, for example regions demar-cated by Vindhya mountains. The data spread seemsto match the orography of the country.
Palghat Gap — Extreme winds observed during mon-soon season over the southern peninsula due to thefunneling effect of Palghat gap (reported wind speedsbeing up to 160 km/h) has been considered.
500t I
Cyclonic Storms — Cyclonic storms on east and westcoasts are more intense over the sea than on the land.West coast storms are less intense except in NorthernSourashtra compared to east coast storms. Reports offailures of structures at Sriharikota, Gujarat coast,Paradeep and other places have been taken into ac-count for evolving the reference speed in these regions.Storm speeds begin to decrease once they cross thecoast and are normally taken to be effective up to adistance of about 60 km inland after striking the coast.However, in Bengal coast its effect has been observedup to 110 km inland. A region along the sea coastextending up to a distance of 60 km (and 110 km alongBengal coast) from the nearest coast line has beenidentified separately from the main land for windzoning. In addition, it has also been decided to give thenumber of cyclones that have struck different sectionsof east and west coasts in the basic wind map for moresophisticated design load estimation.
Dust Storms — Dust storms associated with high localwind speeds which are short lived and are particularlyintense over Rajasthan plains have been considered.Also Norwesters or Kal Baisaki winds over NorthEastern plains during summer months have been ac-counted for. It has been suggested that the influence ofwind speed off the coast up to a distance of about200 km maybe taken as 1.15 times the value on thenearest coast in the absence of definite wind data [5.5]as in Australia. The occurrence of a tornado is possiblein virtually any part of India. They are particularlymore severe in Northern parts of India. The recordednumber of these tornadoes is too small to assign anyfrequency. The devastation caused by a tornado is dueto exceptionally high winds about its periphery and thesudden reduction in atmospheric pressure at its centreresulting in a huge outward pressure on the elementsof the structure. The regional basic wind speeds do notinclude any allowance for tornadoes directly. It is notthe usual practice to allow for the effect of tornadoesunless special requirements are called for as in the caseof important structures such as satellite communica-
++%%
‘.—
“%>_ .—..
!p“-.w,..~
b..
~’.
‘,.
,:
?
-:
,..,’
OPENSEA OPENTERRAIN FORESTStJWRRS CITYCENTFES
FIG. 8 VELOCITYPROFILEOVERDIFFERENTTERRAINCATEGORIES
12
... 9 A-
tion towers. The wind speed can now be obtained upto 500 m height above the ground surface. Design windspeed up to 10 m height from mean ground level istaken to be constant. Although not stated in the Code,turbulence level above 500 m is unlikely to be morethan 6 percent. Basic wind speed map of India has beenpresented in the Code. The reference wind speed allover the country has been found to fall into six rangesof basic wind speeds, varying from 33 m/s to 55 ndsfor a 50 year return period and a probability of ex-ceedence level of 63 percent. The indicated wind speedcharacteristics are in line with other InternationalCodes, namely short duration gusts averaged over atime interval of about 3 seconds applicable to 10 mheight in ‘open’ terrain. The Code recognizes thatthere is not enough meteorological data to drawisotachs (lines joining places of equal velocity) on thewind map of India as seems to be the practice in someother International Codes.
The revised map, when compared to that of the earlierCode shows that in general, in the following areas ofthe country, the wind velocity was previously under-estimated:
— Nollh : North-eastern part of Jammu andKashmir
— West: Coastal areas of Kutch in Gujarat— East: Tripura and Mizoram
The wind map indicates updated supplementary infor-mation (1877 to 1982) relating to the total number ofcyclonic storms. Occasionally, a user comes across asituation where the wind speed at a location on theboundary between two zones has to be obtained. Iflocal wind data is not available, it is recommended thatthe higher speed of the adjacent zones be used fordesign.
3.2 Design Wind Speed [5.3]
The design wind speed is dependent on thegeographical basic wind speed, return period, heightabove ground, structure size and local topography. Theformat of the equation of design wind speed adoptedin the new Code is as follows:
Vz = Vb k, kz k~ ...(6)
whereV. = design wind speed at height z in mls,Vh = basic wind speed of the wind zone,k, = risk coefficient,kz = terrain, height and structure size factor, andk~ = local topography factor.
3.3 Risk Coefficient [5.3.1]
There is sometimes confusion as to the meaning ofreturn period and risk level. Return period (Tin years)is the reciprocal of the probability that a given windspeed will be exceeded at least once in any one year.The word, ‘period’ sometimes causes confusion. ‘Risk
SP64(S&T) :2001
level’, on the other hand means the probability that agiven wind speed will be exceeded in a certain numberof successive years which is usually the life of thestructure. The coefficient kl is identified in the Codefor return periods of 5,25,50 and 100 years alongwithrecommendations regarding ‘mean probable designlife of structures’ of various types [Table 1]. Forgeneral buildings and structures of permanent nature,the return periods have been recommended to be 50years and a life of 50 years for the structure. For alltemporary structures including temporary boundarywalls, it is recommended that a life of 5 years beassumed and the appropriate multiplier kl as given in[Table 1] be used. The normal value of risk levelrecommended is 0.63 for structures whose lifeexpectancy is equal to the return period. The factor klhas been used to assess the effect of different risklevels.
At first sight, a risk level of 0.63, which is the prob-ability that the design wind speed is exceeded once inthe life of say N years of the structure under thecondition that the probability of the exceedence of thedesign wind in any one year is l/N, seems rather high.However experience in many parts of the world hasshown that this amount of risk can be taken for normalstructures. Although not given in the Code, the risklevel can be found from the formula
r=l–(1–l/T)~ ...(7)
whereN = desired life of the structure in years,T = return period in years, andr = risk level.
The designer or owner must decide the amount of risk,he is prepared to take. Let us take one example.
Example. Only 10% risk acceptable, life 30
years, in zone with wind speed of 39 rids,Categoq 2, size of structure 30 m. To find the designwind speed.
From Eqn (7), we get
UT = 1-( l-r)l’N= 3.5x 10-3
That is, this is the acceptable probability that theregional wind speed of 39 rmlswill be exceeded in anyone year during the 30 year life of the structure for theaccepted risk. Using the above values of risk level, lifeof structure and the quantities A and B correspondingto 39 m/s wind speed in [Table 1], one obtainskl = 1.165.
Hence, the regional basic wind speed to be used forthis structure should be 1.165 x 39 = 45.44 rids. Useof Eqn. (7) is necessary if the return period T only isknown from analysis of wind data and risk level has tobe assessed.
13
/
SP 64 (S & T) :2001
3.4 Terrain, Height and Structure Size (kz factor)[5.3.2]
The coefficient k2 has been indicated at differentheights in a convenient tabular form [Table 2] whichidentifies four terrain categories (1, 2,3 or 4) and threeclasses (that is, sizes) of structures (A, B or C). The k2factors are valid only up to the height at which they arespecified. For example, the factor 1.17 in TC-2, classA structure applies only up to 50 m height. As statedin [Note 2 of Table 2], values between 1.12 and 1.17may be linearly interpolated between heights of 40 mand 50 m or a constant value of 1.17 (the higher one)may be used between 40 m and 50 m. The coefficientk2 has different values [Table 33] when dealing withwind speeds averaged over 1 hour, which are requiredfor evaluating dynamic effects according to certaintheories.
3.5 Fetch and Developed Height Relationship[5.3.2.4]
The revised Code had added the concept of fetchlength and developed height for considering the effectof change of terrain category. In the followingexamples the methodology for considering this effectis explained with the help of illustrations.
Example 1. A framed tower structure inside a bigtown at a distance of 0.5 km from the edge of the town;the terrain outside the town isJat open, the structuresinside the town are closely spaced and of heights up to
10 m. Height of the tower 55 m. Tofind the design windspeeds.
Solution: Here, one can take the terrain outside thetown as of category 1 (TC- 1) and the terrain inside thetown to be of category 3 (TC-3). This problem is a caseof wind moving from a lower terrain catego~ to ahigher terrain category. Figure 9 gives a sketch of theconfiguration.
In the above sketch,AA is the edge of TC-1profile (say,where the factor as per [Table 2] is 1.35), B1B2is theapproach terrain of TC- 1 and B2B3is the town centreterrain of TC-3 [5.3.2.1(c)]. The tower is located at C,a distance of 0.5 km from the beginning of the town atB2. In order to apply [5.3.2.3], one proceeds as follows:
At C draw the profile of multipliers as per [Table 2,Class C] (since the height of the structure is more than50 m), corresponding to both the terrain categories 1and 3. Thus CDH corresponds to TC- 1 and CEFG toTC-3. [5.3.2.4] allows one to use two options. As per[5.3.2.4 (b) (i)], one can use the multipliers of thelowest terrain category, which in this case is TC-1, thatis, CDHG. The second possibility is to use the proce-dure of [Appendix B]. To use the procedure of[Appendix B], one first notes the height up to whichthe TC-3 profile has penetrated in the vertical directionin the terrain in which the structure is located. In thiscase, [Table 3] indicates that TC-3 has penetrated up
,/
L ..-.__.._,.,
to a height of 35 m at a distance of 0.5 km from B2. ‘aI $Hence one takes the multiplier of TC-3 [Class C of ,,
Table 2] up to a height of 35 m (by interpolation).
t--i
-.—
Above this height, the multipliers of TC-1 are to be *
used. At a height of 35 m, the multiplier correspondingto TC-3 is to be used when computing the design load
I
...
on the tower, although the multipliers of both TC- 1and tTC-3 are shown at this height in this sketch. The finaldesign multipliers are shown in Table 3.
Table 3 Multipliers for a Structure in TC-3 withUpstream TC-1
@ample 1)
Height (m) Multipliers as per Multipliers as per[5.3.2.4(b)(i)] [AppendixB]
10 0.99 0.82
15 1.03 0.87
20 1.06 0.91
30 1.09 0.96
35 1.10 1.04
50 1.14 1.14
55 1.15 1.15
NOTE- It is observedthat the applicationof [5.3.2.4(b)(i)]leadsto higheror conservativeload.
Example 2. A chimney of height 150 m in opencountry at a distance of 10 km from clusters of scat-teredsmall villages all round with houses of height lessthan 10 m. Tofind the wind speed multiplier profile.
Solution: Since the terrain up to a distance of 10.0 kmfrom the chimney is open, one may assume that theterrain is of category 1 up to 10 km from the chimney.In view of the location of clusters of small villagesbeyond 10 km, one can consider the approachingprofile to chimney to be TC-2 from beyond 10.0 kmand TC- 1 for a distance of 10 km from the chimney.Figure 10 gives the sketch of the configuration. It maybe observed that in this problem, the wind flows froma higher terrain category to a lower terrain category. InFig. 10,AAis the edge of the TC-2 profile (say the lineof multiplier 1.35), B1B2is the approach terrain TC-2and B2B3the TC- 1 in which the chimney is located.The chimney is located at C, which is located at adistance of 10 km from the beginning of TC- 1,namelyB2. To obtain the appropriate multipliers, one canproceed as follows:
i) Assume that TC- 1multiplier applies for theentire height, as per [5.3.2.4 (b) (i)]
ii) Use the procedure of [Appendix B] as follows
To use the procedure of [Appendix B], draw the varia-tion of the multipliers of TC-2 and TC-1 at C withheight. These are mpectively CEFG and CDH. From~able 31of the Code,it is observed that the profile ofTC-1 penetrates up to a height of 80 m, in a distance
.
of 10 km. Thus one takes the TC- 1 multipliers of ClassC (since the maximum chimney dimension exceeds 50m) up to a height of 80 m. This multiplier has aninterpolated value of 1.17 at a height of 80 m. Abovethis height, one has to take the multipliers as per TC-2,Class C. But the multiplier as per TC-2, Class C has avalue of 1.17 only at a height of 100 m as per [Table2]. The recommendation of [Appendix B] is to take thesame value of the multiplier as per the terrain oflocation (that is TC-1) up to the height of the upstreamhigher terrain category at which the same multiplier isfound as per [Table 2] of the Code. In this case, therecommendation implies that the multiplier 1.17 at 80m height of TC- 1 must be continued up to a height of100 m, above which only the TC-2 multipliers of ClassC are to be used. Table 4 gives the final recommendedmultipliers.
SP64(S&T):2001
Table 4 Multipliers for a Structure in TC-1 withUpstream TC-2
(Example 2)
Height (m) Multipliersas per Mukiptieras per[5.3.2.4(b) (i)] [AppendixB]
10 0.99 0.99
15 1.03 1.03
20 1.06 1.06
30 1.09 1.09
I 50 I 1.14 I 1.14 I
I 80 I 1,17 I 1.17 I100 1.20 1.17
150 1.24 1.21
NOTE—Asobservedearlier[5.3.2.4(b)(i)]giveshigherloadsbut is simplerto apply.
A — . .— . -+-—–A
—.
v
+
/
/
+
TC1
—.— Tc .1
~ TC-3
— DESIGN PROFILE
B3
FIG.9 DETERMINATIONOFVELOCITYPROFILENEARACHANGEINTERRAINCATEGORY(LESSROUGHTOMOREROUGH)
.- .-
p——- -----
I 1
,’
I
15
SP64(S&T) :2001
A .— . .— .— -—. .— A
TC2
—..
..
./
Tcl
BI IB2
tii’7,xiii#/
/’
Ic 83
~— TC-1 PROPILE--—- TC-2 PROPILE
DEStGNPWFtLE
FIG. 10 DETERMINATIONOFVELOCITYPROFILENEARCHANGEINTERRAINCATEGORY(MOREROUGHTOLESSROUGH)
Example 3. A TV tower of height 300 m located in the
centre of a city with tall buildings (greater than 50 min height) up to a height of 3 km from it, closely builthouses of up to 10 m height from 3 km to 10 km andjlat terrain beyond for 15 km at which sea coast is
encountered. Wind is from the sea. To find the windspeed multipliers.
Solution: This k a problem in which wind passesthrough four terrains before reaching the structure ofinterest. At high winds, the sea can be taken asequivalent to TC-2 due to the large waves that aregenerated. In this problem, TC-2 is followed by TC-1for 15 km and this is followed bf TC-3 for 10 km witha final terrain category 4 for 3 km. The configurationis sketched in Fig. 11 where the multiplier profiles ofTC- 1 to TC-4 are shown at the location of the TVtower. As before, one can choose the simpler methodof [5.3.2.4 (b) (i)] and use the multipliers of TC- 1 forthe entire height. To use the alternate method given in[Appendix B], one draws first the developing profilesfrom the beginning of each terrain [Table 3].
a) AA is the edge of TC-2 (say the line of multi-plier 1.35).
b)
c)
d)
e)
B1B2 is the approach terrain TC-2, B2B3is thefirst intermediate terrain TC- 1, B3B4 is thesecond intermediate terrain TC-3 and B4B5isthe final terrain TC-4 in which the TV tower islocated.From Vable 3], it may be easily determinedthat the edge of the TC- 1 profile starting at B2meets the TV tower at“Clat a height of 100 m,the edge of TC-3 profile starting at B3meets theTV tower C2at a height of 250 m and the edgeof the TC-4 profile starting at B4meets the TVtower at C3 at a height of 220 m and the edgeof the upstream TC-2 profile is at a height of490 m,All the four category profiles (of class C sincethe maximum dimension of the structure ismore than 50 m) have been drawn with originat C, the base of the TV tower.Proceeding as before, one finds that TC-4swamps TC-1 and is effective up to a height of220 m, from C to D. At D, one”encounters TC-3beginning at 1$ up to a height of 250 m. Thusthe profile shifts from D to Eon TC-3 profileat a height of 220 m and moves along the TC-3profile from E to F, which is at a height of
,,
/1
I
16
I,, I
1!
A— Jui?- —
sP64(s&T):2001
.—— —— — .— —— —_ _
ITH
-A
TC-2 lc-tSkm 1b3 7 km B,~ 3km “
, 7 1
— DESIGN PROFILE ‘A— TC-1 PRCFHE —x— TC-3 PROFILE—o— TC-2 PROFILE ----— Tc-b PROFILE
FIG. 11 DETERMINATIONOFDESIGNPROFILEINVOLVINGMORETHANONECHANGEINTERRAINCATEGORY
250 m. Above F, the unpenetrated profile is thefar upstream TC-2 profile and this applies forthe rest of the height of the TV tower. Table 5gives the final multipliers.
Table 5 Multipliers for a Structure in TC-4 withUpstream TC-3, TC-1 and TC-2
(Example 3)
Height (m) Multipliers as per Multipliers as per[5.3.2.4(b)(i)] [AppendixB]
10 0.99 0.67
15 1.03 0.6720 1.06 0.8330 1.09 0.8350 1.14 1.05100 1.20 1.10150 1.24 1.10200 1.26 1.13220(intetpulated) 1.27 1.19 (TC-3 vatue)
250 1.28 1.26 (lW2 value)
300 1.30 1.31(TC-2value)
3.6 Topography (kJ [5.3.3]
The coefficient k~allows for undulations in the localterrain in the form of hills, valleys, cliffs, escarpmentsand caters to both upwind and downwindslopes. Methods for evaluation of k3 are given in[Appendix C] of the Code.
17
3.7 Design Whd Pressure [5.4]
One of the important modifications in the Code relatesto the veloeity-pressure relationship. In the earlierversion of the Code, the pressure was over-estimatedby about 25 percent for a given wind velocity. Thecorrect equation given now reads as follows:
pz = 0.6 VZ2 ...(8)
where
Pz = design wind pressure at height z in N/m2, andVZ = design wind speed at height z in mh.
Strictly speaking, the coefficient 0.6 is only the mostprobable average for the atmospheric conditionsprevailing in India during the whole year. A value of0.6125 would be more appropriate in temperate zonesabove latitude of about 33°. The wind effects in termsof static loading on the structure are determined fromdesign pressures as indicated in 4. A comparison ofwind pressures as per new and 1964 Code is given inthe following example.
Example. Estimation of design wind pressure for abuilding as per 1S 875:1964 and IS 875 (Part 3): 1987.
Given:A buildingof length35.0q width 15.0m, height15.0mLife of structure 25 yearsTermincategory 3Location:Delhi
SP 64 (S & T) :2001
Topography: Almost flat
Solution:
Previous code [1S875: 1964], [Fig. 1A] gives pz=150kgf/m2.
Present code [1S875 (Part 3) : 1987] :
Design wind speed VZ = Vb.kl.kz.ks
Vb = 47.0 m/s [Fig. 1]
kl = 0.90 [Table 1, N = 25 years]
Since greatest horizontal dimension of 35 m is between20 m and 50 m
k2 = 0.88 [Table 2, Class B] at 10.0 m height
k~ = 1.00
Hence Vlo = 47.0 x 0.90x 0.88x 1.0= 37.22 nds
Design pressure at 10 m height = 0.6 V;
= 831.37 N/m2= 84.83 kgf/m2
(g= 9.80 m/s2 in most parts of India)
Let the life of the structure be increased to 50 years[kl=l .0, Table 1], then
plo = 0.6 [47.0x 1.0x 0.88x 1.0]2 = 1026.39 N/m2= 104.73 kgf/m2
If the structure is located in a terrain category 2,
then the value of k2 = 0.98 andPll) = 0.6 [(47.0)(1.00)(0.98)(1.00)]2
= 1272.91 N/m2= 129.89 kgf/m2
Let us now consider a situation where the same struc-ture is situated at Madras, Calcutta, Roorkee, Bombay,Bangalore and Darbhanga. The design wind pressurewill be as given in Table 6, for different possible terrainconditions. From Table 6, it is observed that the designwind pressure is strongly affected by the considerationof the expected life of the structure and terraincategory. It is observed that the earlier Code generallyover estimated the wind load in city centres. One canalso work out that the earlier Code generally underes-
, timated the design wind load, if either the expected lifeor ‘return period’ is large (for example, seethe case forBombay in Table 6). This is particularly true forregions where the design pressures, according to the1964 Code are low.
4 WIND PRESSURE AND FORCES ONBUILDINGS/STRUCTURES [6]
4.1 General
The Code stipulates requirements for calculation ofwind loading from three different points of viewf
Table 6 Design Wind Pressure(Example )
Location
“e “’rain m(Yeara)
1987 1964
25 3 831.37II
DelhiI
50 3 1026.39 1470.00
50 2 1271.91
Calcutta 25 3 940.89 1962.00
Roorkee 25 2 741.84 1471.50
25 3Bombay
744.91981.00
50 2 1115.60
Madraa 25 3 940.89 1962.00Bangalore 25 3 447.09 981.00
Darbhanga 25 2 1380.73 1471.50
— The building/structure taken as a whole;
— Individual structural elements such as roofsand walls; and
— Individual cladding units such as sheeting andglazing including their fixtures.
The wind loading is given in terms of pressure coeffi-cients CP and force coefficients Cf and can be deter-mined as follows:
F= CfAep~ ...(9) [6.3]
F= (Cw _ Cpi) A Pd ...(10) [6.2.1]
where
F=
Cf =
A. =
pd =
Cpe*Cpi =
A=
wind load;
force coefficient;
effective frontal area obstructing wind,which is identified for each structure;
design wind pressure;
external and internal pressure coeffi-cients; and
surface area of structural elements.
4.2 Pressure Coeftlcients [6.2]
Pressure coefficients are applicable to structuralelements like walls and roofs as well as to the designof cladding. The calculation process implies thealgebraic addition of Cw and C~i to obtain the finalwind loading by the use of Eqn. (10). The Codeindicates both these coefficients separately for a widevariety of situations generally encountered in practice.
4.2.1 External Pressure Coefficient CF [6.2.21
External pressure coefficient depends on wind direc-tion, structure configuration in plan, its height versus
18
I
.’
width ratio and, characteristics of roof and its shape.[Tables 4 to 22] are devoted exclusively to the deter-mination of CPVFor the specific design considerationsrelating to cladding, local coefficients have beenseparately shown delineating the areas at the edges ofwalls and roofs where high concentration of negativepressure is often found to exist.
4.2.1.1 External pressures
It deals with a large number of cases of gross forcecoefficients and pressures. Should pmssums in[Table 4] be applied at each level in the case of a tallbuilding, with velocity variation as in [Table 2]. Theanswer, is that the Code implies that the calculationshould be carried out for each level as if the air load atthat level is independent of that at an adjacent (or loweror higher) level. It is puzzling to note a positive pres-sure coefficient of 1.25 in [6.2.2.T, when one knowsthat it cannot exceed +1.0. The reason is that since thepressure coefficients on the small overhanging posi-tions are not given but are known to be more stronglynegative on top than the negative pressures on the non-overhanging portions, these numbers are expected tocompensate for the projected lower pressures on thetop of overhanging portion. Although not stated in theCode, it is desirable to take the same values in [6.2.2.Ton the vertical walls above and below the overhangingportion, over a height equal to half the projectinglength of the overhang; if no other guideline is avail-able.
4.2.2 Internal Pressure Coefficient CPi[6.2.3]
Internal pressure coefficients are largely dependent onthe percentage of openings in the walls and theirlocation with reference to wind direction. The Codeindicates C~i for a range of values with a possiblemaximum (that is positive pressure) and a possibleminimum (that is negative pressure) with the provisionthat both the extreme values would have to beexamined to evaluate critical loading on the concernedmember. Three cases have been specifically indicatedfor arriving at C~i:
—
—
—
openings up to 5 percent of wall area,
openings from 5 percent to 20 percent of wallarea, and
openings larger than 20 percent of wall area(including buildings with one side open).
The last case is of particular interest in determiningwind loading for structures such as aircraft hangerswhich have wide full height openable doors formingone side of the enclosure [Fig. 3].
sP64(s&T):2001
4.2.2.1 Internal pressures [6.2.3]
The internal pressures given in [6.2.3] have beendeliberately specified slightly higher (negatively) thanthe codes of temperate countries to reflect the fact thatas a tropical country, the size of windows/doors arelarger in India and the normal tendency is to keep themopen as much as possible.
4.3 Force Coefficients [6.3]
Force coefficients applicable to the buildinghmctureas a whole as well as to structural frameworks whichare temporarily or permanently unclad are also givenin [6]. For evaluating force coeftlcients for the cladbuildinghtmcture as a whole, the Code gives guidancefor a variety of plan shapes and height to breadth ratios[Table 23]. It also indicates extra forces occurring dueto ‘frictional drag’ on the walls and roofs of cladbuildings [6.3.1]. The diameter D to be considered forrough surface when applying [6S.2.2] is the valueexcluding the height of the roughness, that is excludingthe height of flutings or similar ‘regular’ roughnessesor excluding average height of random roughness likesand particles. This definition of D is valid only ifD/E >100 (where Eis the average height of the surfaceroughness). For still larger roughnesses, specialistsadvice should be sought.
Force coefficients on unclad buildingshtructures,frameworks and their individual members, are com-prehensively covered in [6.3.3.3 to 6.3.3.6]. Theframeworks included are those that are single (that isisolated), multiple or in the form of lattice towers. Theeffects of ‘shielding’ in parallel multiple frames andthe effect of different solidity ratios have been incor-porated along with global force coefficients for latticetowers Uables 28 to 32]. Force coetllcients for in-dividual members of various structural shapes andwires/cables have been separately indicated [Tables 26and 27].
4.3.1 Individual Members [633.2]
The force coefficient of rectangular members given in[Fig. 4] reflect the recent discovery that the aspect ratioof the nxtangle affects the total force coefficient andpeaks at an aspect ratio of about 2/3. Slight discrepan-cies will be observed between the gross value in[Fig. 4] and the summed up values in [Table 4], forsmaller height to width ratios and the value for thesquare in [Table 26]. Such seeming contradictions willbe observed in several international codes also andreflect, mainly the continuation of values taken fromearlier codes. The last case in Pable 4], has reconciledthe value in [Fig. 4] and that in the last sketch of[Table 4], hlw26.
?.- .—— -:,-.
I
$
,*,
?
i
/
19
I
SP64(S&T):2001
In the other cases it is recommended that the negativepressure behind the body be further reduced to bringthe total force to the value in [Fig. 4]. For example letus consider the case 1/2 c hAvS 312,312< Uw c 4 and8 = 0° in [Table 4 (item 4)] with a = 2, b = 8 and h= 3. The total force in the direction of wind withpressure coefficient of +0.7 on faceA and -0.3 on faceB is 1.0. ‘a’ in [Fig. 4] is ‘w’ of [Table 4], ‘b’ in [Fig.4] is ’1’of [Table 4] and ‘h’ has the same meaning inboth. Hence
hJw= Ida = [h/b. blaJ
Therefore,
hlb = hlw. alb
llw = bla
For hlw = 3/2 and llw = 4 of [Table 4] we have to lookup Wb = 3/8, db = 0.25 in [Fig. 4B] which gives Cf =1.18, while adding the pressures on faces A and B in[Table 4] yields Cf = 1.0. The recommendation is notto change the pressure on the front face A (they areusually more reliable and in any case cannot exceed1.0), but decrease the pressure on the back face from-0.3 to -0.48. The pressures on the sides as well as atthe corners need not be changed.
4.3.2 Force Coefficient for Framed Structures[6.3.3.3 and 6.3.3.4]
The force coefficients for framed structures given in[6.3.3.3] and [6.3.3.4] reflect the result of data whichbecame available around 1980. A few designers haveencountered a condition (in unusual structures), wherethey found that the sum of the total force on more thanthree frames was more than that on a solid body of thesame outer geometry. This is correct and has to beapplied as such if all the frames are finally mounted ona single large structure. In some cases, the designershave found it more economical to have a fully cladstructure with an internal braced framework.
5 DYNAMIC EFFECTS [7]
The Code incorporates a new section relating todynamic effects, recognizing the advent of a largenumber of tall, flexible and slender structures in thecountry’s sky-line. Structures which require investiga-tion under this section of the Code are the following:
— those with height to minimum lateral dimen-sion ratio of more than 5.0.
— structures with a first mode natural frequencyof less than 1.0 Hz.
Guidelines are given in the Code for the approximatedetermination of natural frequency of multistoreyedbuildings and the designer is cautioned regarding cer-tain structural responses such as cross-wind motions,interferences of upwind obstructions, galloping, flutterand ovalling. The Code encourages use of wind tunnelmodel testing, analytical tools and reference to
specialist advice when wind induced oscillations of thestructure reach significant proportions. [Clause 7.1]gives guidelines for the assessment of dynamic effectsof wind. It is to be noted that the Strouhal number fornon-circular bodies [7.2] is usually slightly less thanthat for circular bodies.
6 GUST FACTOR (GF) OR GUSTEFFECTIVENESS FACTOR (GEF) METHOD [8]
6.1 One doubt that may arise during a comparativestudy of [Table 33] to find hourly mean wind speed,and peak gust multiplier from nable 2] is the absenceof the class of structure that is its size in [Table 33].However, further study will show that the variable $takes into account the building size [8.3 and Fig. 8].
The Code says that the use of random response methodfor across wind response of structures of non-circularcross-sections have not matured and hence are notgiven in the Code. Some users may find small dif-ferences in the graph for E [Fig. 11] and those in someother codes. Because of the local short period intensenature of wind in most parts of our country, the calcu-lated gust factors as per the Indian Code will be foundto be generally slightly higher than in other codeswhich are based on the turbulence characteristics ofthe ‘steady’ fully developed pressure system wind.
6.2 A1ong-Wind Loads [8.3]
The use of gust factor approach has been permitted bythe Code for the evaluation of along-wind load.
The calculations involved for the estimation of along-wind load have been reduced to a simple equationgiven below:
FZ=Cf. AepZ~ G ...(11) [8.3]
whereF, = along-wind static loading at height z, applied
on effective frontal area Ae of strip underconsideration,
Pz’ = design pressure at height z corresponding tohourly wind speed, and
G = gust factor.
One of the important departures for along-winddynamic effects as compared to static effects is toapproach the problem of design pressure evaluationfrom the route of hourly wind instead of short durationgusts as one is required to do for application up to [6].[Table 33] indicates the revised factors k2which haveto be applied to the basic wind speed (short durationgusts) of [Fig. 1] for arriving at the hourly mean windspeed and consequently the corresponding designpressures, pz’.
The gust factor G, which depends on natural frequen-cy, size and damping of the structure and on the windcharacteristics can be evaluated directly fromparametem obtained from graphs given in [Fig. 8 to 11].
,’
20
,.
6.2.1 Computation of Wind Load by Gust FactorMethod
Step 1. Hourly Mean Wind (V;): Hourly mean windspeed is maximum wind speeds averaged over onehour.
VZ’= V~.kl .kz’.k~ ...(12)
wherev~ = Regional basic windspeed[Fig. 1],k, = Probability factor [5.3.1],kz’ = Terrain and height factor [Table 33], andor k2k~ = Topography factor [5.3.3].
Step 2. Design Wind Pressure (p;) [6.2.1]
p;= 0.6 (V~)2 ...(13)
Step 3. Along-wind Load [8.3]
F== @ie.p:.G ...(14)
r
I
whereFZ =Cf =A. =
Pz’ =G=
along-wind load on structure at any height Gforce coefficienteffective frontal area;design wind pressure;the Gust Factor = (peak load/mean load); andis given by
G=l+gf. r{[B(l+@)2+ SE@] ...(15)
where
~f =
r =
B=
SEl~ =
s =E=
6=
peak factor defined as the ratio of the ex-pected peak value to the root mean value of afluctuating load;roughness factor;gf. r is given by [Fig. 8]background factor [Fig. 9];measure of the resonant component of thefluctuating wind load;size reduction factor [Fig. 10];measure of available energy in the windstream at the natural frequency of the struc-ture [Fig. 11];damping coefficient of the structure[Table 34]; and
g~. r.ti(+= ~
and is to be accounted only for buildings lessthan 75 m high in terrain category 4 and forbuildings less than 25 m high in terraincategory 3, and is to be taken as zero in allother cases.
6.3 Across-Wind Loads
The Code points out that the method of evaluatingacross-wind and other components of structuralresponse to wind loading are fairly complex and havenot as yet reached a stage where successful codifica-
SP64(S8ZT):2001
tion can be attempted. It recommends use of windtunnel model studies and specialist advice in caseswhere significant across-wind loads are anticipated.
7 WIND TUNNEL MODEL STUDIES [1 and 7]
The Code recommends that model studies in wind tun-nels lx resorted to under several conditions. These are:
i)
ii)
iii)
for structures of unconventional shapes, un-usual locations and abnormal environmentalconditions, not covered in the Code [1.1.3,Note 1];certain wind induced dynamic responseproblems [7.1, Notes 2 and 9]; andbuildings or shapes for which more accuratedata or conflation of existing data is required.
Unlike in some foreign codes, no guidelines have beenprescribed for properly carrying out such studies orlimitation of model studies including those due toReynolds number etc. The mason appears to be thatthere are very few suitable wind tunnels in our country.However, since the existing ones are being improvedand four to five new wind tunnels are likely to comeup in the next 3 to 5 years, it is in order to discuss this.
It is now well recognized that wind load on a structure(static or dynamic) depends on:
i)ii)
iii)
Variations of mean velocity Uo,with height;Variation of the intensity of turbulence
*
with hei ht, the intensity being defined as
where u’, v’ and w’ are the
departure of the velocity at any time along x,yand z axes, respectively, from Vo, andthe ‘scale’ of turbulence, which is a measure ofthe size of coherent blobs of flow, usuallyswirls of various sizes, called ‘eddies’.
The above is not an exhaustive list, but is generallyadequate for consideration. Creation of the meanvelocity distribution in a wind tunnel is a relativelysimple task and about half a dozen methods are avail-able for doing it. However, care has to be exercisedwith regard to the simulation of (ii) and (iii). Some-times, it is stated that the intensity of distribution ofturbulence and scale must be the same as values scaleddown from that in the atmosphere. This maybe accept-able for sharp edged bodies whose properties do notdepend on the Reynolds number. But the force coeffi-cients on rounded surface bodies depend strongly onReynolds number and while the spectra must be simu-lated, the intensities may have to be different, to ensurethat force coefficients are not measured in the forcebucket (see data for R, between 4 x 105 and 2 x 106 in[Fig. 5] which shows the ‘buck~t’).
The above discussions lead to the followingguidelines for wind tunnel model studies.
general
21
/
SP 64 (S & T) :2001
i)
ii)
iii)
iv)
The variation of the mean velocity with heighthas to be modelled, to within 2.0 percent to3.0 percent.The longitudinal component of turbulence in-tensity, particularly, has to be modelled suitab-ly to ensure that the non-dimensional spectralenergy n.S.(n)/ cr. has the same variation withrespect to n.L.#VZ as in the atmosphere, espe-cially around the natural frequency (or fre-quencies) of the structure.The model dimension along the wind shouldnot exceed three times the longitudinal scale ofturbulence. This requirement helps in fulfilling(ii) to some extent.All scaling laws used to extrapolate modelresults to full scale, incorporate establishedReynolds number effect correlations and suchcorrections as may seem necessary should beapplied. It is worth stating that the instrumenta-tion chain used must have adequate responsecharacteristics.
8 APPLICATION OF CODAL PROVISIONS
8.1 Provisions of IS 875: 1987 can be broadlyclassified as:
a)
b)c)
d)
e)
Computation of design wind speed based onwind zone, terrain category, topography andwind direction.Computation of design wind pressure.Computation of wind load using pressureco-efficients.Computation of wind load using force coeffi-cients.Computation of along-wind force using gustfactor method.
8.2 For easy understanding, the above steps of com-putation are detailed in the form of flow charts inFig. 12 to 15. Reference to clauses, tables and chartsof IS 875 (Part 3): 1987 has also been incorporated inthese flow diagrams. Further explanation of the abovepoints is given alongwith the solved examples.
,,
I “
22
i:~1’,
SP 64 (S & T) :2001
&
MEAN PROBABLE LIFE
●
I FIND Vb FOR PARTICULAR ZONE[REFER FIG.1 OR APPENDIX Al
-REFER [CIAUSE 5.3.2,3] FOR EFFECTS OFCHANGE IN TERPINN CATEGORY DUE TOFETCH & DEVELOPED HEIGHT
1●
-Yesl+
No 1
Yes
I CALCULATE K3 AS PER
[APPENDIX C] II
Ai
Jw
iNo
CALCULATE
VZ= Vb, KI, K2, K3 I*CALCULATE GESIGN WIND
PRESSUREPd = 0.6 VZ2 AT DIFFERENT HEIGHTS
oSTOP
..
. ...—
\
$
FIG. 12 FLOWDIAGRAMFORCALCULATINGDESIGNWINDPRESSURE
23
/
SP 64 (S & T) :2001
I CALCULATEWIND PRESSUREPd = 0.6 VZ2 I
TO READ COEFFICIENT OF PRESSURE CP & TO CALCULATEWIND LOAD NORMAL TO A SURFACE FOLLOW
STEPS BELWJ
(fj +.+‘ REFER CIAUSE 16.2.33 & u
“ ITABLE 71::O~~OSLWE &FREE.~/Q3LE 8]: FC4?FREE STANDIW
DOUBLE SLOPED R~FS* ~ABLE 9] ; FOR PITCHED FREE ROOF
&a=30”* ~ABLE 101: FOR PITCHED FREE ROOF
&a. 3& WITN EFFECT OFTRAIN OR STORED MATERIAL
● ~ABLE 11]: FOR PITCHED FREE ROOFS
&a=lO”“ ~ABLE121:FOR PITCHED FREE ROOFS
& u = IO” WITH EFFECT OFTRAIN OR STORED MATERIAL
‘ ~ABLE 13]: FOR TROUGHED FREE ROOFS
aci= lo”‘ ~A13LE 14]: FOR TROW3HE0 FREE ROOFS
& a = 1~ WITH EFFECT OFTRAIN OR STOREO MATERLAL
1622S6TA~
FOR Cm IF CYLINDER WITH AXIS
a) NORMAL TO GROUNDb) PARALLELTO GROUNOc) ~ > Io,ooo
FOR CX = 0.S IF MD ~ 0.3= 0.5 IF MD< 0.3
Nets: h Is tiw heiiht 84a uerlicalcylinderor bngth of a horizontalcylinder,If there is a fmeSowofair aroundbdh ends, h b to betaken aa halftha lengthfor mmpu-tlngthe rallo of MD.
“Km##mwuFOR Cm C+ ROOFS & SOITOMS OFCYUNORICAL ELEVATEDSTRUCTURES
Yes
U!lfiStNEO
ROOFS & ROOFS WITHA SKYLIGHT
REFER CIAUSF
19J2.116TASLF2 11
FOR A TYPICAL GRANDSTAND PROOF
‘~FOR UPPER SURFACE OFROUND SILOS & TANKS
FIG. 13 FLOWDIAGRAM(2)TOREADCPVALUESOFBUILDING/STRUCTURES(Continued),.
24
I
6A
IS IT NoSINGLE SPAN
SP 64 (S & T) :2001
I ‘“s47“r!YzER. MUSE [6.2.2.6
& TABLE 16FORC- F9RPITCHEDROOFSOF MULTISPANBUILDINGS WITHh NOTGREATERTHANW’
REFERQ&&E 16.2.2.6
E 171FOR Cm FORSAWTOOTl+EDROOFS OFMULTISPANBUILDINGS WITH
,i &&yes,?REFFR Cl AUSE
[6.2.2.1 & TABLE 4]
FOR Cm OFRECTANGULAR CUD
BUILDING
No i No
IsAREA OF OPENING
QU
REFER CLAUSE
BETWEEN 5“h & 20% OF Yes u
WALL AREA FOR C@= A().2
No
h● REFERCIAUSE f6.2.3.2]
AREA OF OPENING>20% OFWALL AREAC*= ko.7
● REFER IFIG.3] FOR BUILDINGWITH ONE SIDE OPENDEPENDING UPON B/L RATIO
REFER CLAUSE r6,2.2.2 & TABLE Q
FOR Cp OF PITCHED ROOFS OFRECTANGULAR CIAD BUILDINGS
REFER CLAUSE f6.2.2,~
FOR OVERHANG ROOFS
REFFR CLAUSE 16.2.2.3 & TABLE 6]
FOR MONOSLOPE ROOFS OFRECTANGULAR CIAD BUILDINGS
REFER CLAUSE [6.2 2.5 & TABLE 15]
FOR C= OF CURVED ROOFS
FIG. 13 now DIAGRAM(2) TOREADCPVwum OFBUILDINGS/STRUCTURES
..
-. ...—._,-
,!
~—-.. -----
i
Wj
11
,
>.
,.
/
25
I
SP 64 (S & T) :2001
oEslaI%&%f%suREP*= o W*2
J●
TOREADF- COEFFICIENTOFmLowwmRucll#lE ~
TNE8ESTEP6
&
+ ‘=
tTA BUHDINQ
+
FOROTHER CLAoBUK.OINGSw
UNIFORM SECTleMOBTAIN ~VNUEFROM fTABLE23]
1& FlG4q
FORqVALUEcORmEsmNOINcToatl mM21 mF* qvALuE
CORRESPONUNGTOmbORNb *1 c)B
..“.,_ ____.
_.—.
f
k“
:..
I\
1
I,
FIG, 14 FLOWDIAGRAM(3) TOREADFORCECOEFFICIENT(Continued)
,,
26
..SP 64 (S & T) :2001
0B
Yes
ANo
1= [T Yes REFtR ITABLE23] i
+
DEPENDlt4G ONDIAMETER (D),DESIGN WIND
SPEED (Vd)&SURFACEROUGHNESS
LREFER CLAUSE[6.3.3.4,TABLE29
& FIG. 7] FOR ~
Yss
rk
-1 I
INo
FORCOMPUTING C,FOR FRAMESHAVINGACOMBINATIONOF CIRCULARAND FLATSIDED MEMBERSUSING INTERACTIONRELATIONSHIP
FIG. 14 FLOWDIAGRAM(3) TOREADFORCECOEFFICIENT
f RLAT71CE
RFFER ITABLE ~FOR SQUARE OR
EQUIU4TEIULTRIANGuLARBASEWITH FIAT SIDEOMEMBERsfZEFERmaFOR SQUARE SAsEWITH ROUNDEDMEMBERs
TRIANGULARBASEWITH ROUNOEDMEMBERS
{
. ..—.–—..
,}$
_ -.—___
1f
!+*’ ;..,,i ‘j
,’.,
J
,,’
I
27
I
‘1I1!
SP 64 (S & T) :2001
( GUST FACTORMETHOD ~
ICALCULATE HOURLYMEAN SPEED (VZ~AT
HEIGHTz AND TERRAINCATEGORYVZ ‘Vb, KI,K2, K3
ICALCULATE DESIGN WIND PRESSURE ATHEIGHT Z DUE TO HOURLY MEAN WIND
SPEED I
I Pz = 0.6.(VZ’)2 IREAD m.r & L(h) FROM (FIG.8)
DEPENDINGON THE HEIGHTOFBUILDING/STRUCTUREAND
TERRAIN CATEGORY
READ”BACKGROUND FACTOR BlFROM FIG.9]
DEPENOINGON CZ.hlL(h) & ~Where, A = Cy,b/Cz.h
Cy= 10, Cz= \2:b IS BREADTH OF STRUCTURE
NORMAL TO WIND STREAM& h IS HEIGHT OF STRUCTURE
I*
1
IREAD SIZEREDUCTION FACTORS
OM IFIG. 10]FOR THE VALUE OF REDUCED I
IFREQUENCY Fo = Cz . fo,h/v’h & k
Where f. IS THE NATURALFREQUENCY OF THE STRUCTURE
!L
I READ GUST EN R YFACTOR E FROM FI .11
FOR THE VALUE OF‘: ‘1I F(I=CZ fr).~”h
I
ICOMPUTE GUST FACTOR G ( = peak load/Mean load) USING THE EXPRESSION
I G=l+gr. r[B(l+@)2+S. E/p]%
Where ~ IS THE DAMPING COEFFICIENT=0.010 FOR WELOED STEEL STRUCTURE=0.020 FOR BOLTED TEEL STRUCTURE=0.016 FOR REINFORCEDSTRUCTURE
oSTOP
FIG. 15 FLOW DIAGRAMFORGUSTFACTOR
28
. - -——-,
:,
,1’
SP64(S&T) :2001
9 ASSESSMENT OF WIND LOADS ON BUILDINGS AND STRUCTURES
9.1 Pressure Coefficient Method
9.1.1 Stepwise computation of wind load using Peak Gust Method is as under
Step 1. Design Wind Speed (VZ) [5.3]
Vz =
where
Vb =
Vz =
k, =
kz =
k~ =
V~.kl.k2.k~
basic wind speed,
design wind speed,
probability factor or risk coefficient [Table 1],
terrain, height and structure size factor [Table 2], and
topography factor [5.3.3].
Step 2. Design Wind Pressure (pZ) [5.4]
P. = 1/2p VZ2
where
P = the density of air and its value is 1.2 kg/m3.
Step 3. Wind Load Calculation [6.2.1]
The wind load on individual structural elements such as roofs, walls, individual cladding units and their fittingsis calculated by accounting the pressure difference between opposite faces of these elements or units and isgiven by:
F=
where
F=
Cpe =
Cpi =
A=
P(j =
(CP - cPi) A.pd
wind load,
external pressure coefficient,
internal pressure coefficient [6.2S.1],
surface area of structural element or cladding, and
design wind pressure.
Values of CPeand C~ifor different types of roofs, walls of different types of buildings are given in [Tables 4 to22].
Step 4. Estimation of Frictional Drag (F’) [63.1]
Depending upon the d/h or d/b ratio of rectangular clad buildings additional drag force given by Eqn. 16 or 17is added to the force given by [6.2].
if hSb F= C~(d–4h) b.pd+C~(d–4h) 2h.pd ...(16)
h > b F = C; (d – 4b) b.p~ + C; (d – 4b) 2h.p~ ...(17)
Values of C~ are given in [6.3.1] for different surfaces.
9.1.2 Solved Examples of Rectangular Clad Buildings
Example 1. A rectangular clad building having pitched roof and located in a farm (Fig. 16) [Tables 4 and 5].
Given:
Physical Parameters:
I
!
29
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SP 64 (S & T) :2001
Height (k) 3.5 mWidth (w) 10.0 m (excluding the overhangs)Length (1) 18.0 mRoof angle (a) 5°Overhang 0.5 mOpenings on sides : 10 percent of wall areaExternal surface of walls : SmoothFlat ground
~
18m
FRONT ELEVATION
3k m
-J_
FIG. 16 A RKTANGULARCLADBUILDINGwrrH PITCHEDRGGF
Wind Data
Wind zone : 3(Basicwindspeed =47m/s)Terrain category : 1 [5.3.2.1]Class of structure : A [5.3.2.2]
Design Wind Speed (VIO)
VZ = V~.kl.k2.k~
V~ = 47 nds [5.2, Appendix A]
kl = 0.90 (Farm Building) [5.3.1, Table 1]
kz = 1.05 [5.3.2.2, Table 2]
k~ = 1.00 [5.3.3.1], for site upwind slope less than 3°
Hence
v 10 = 47x0.9x 1.05X 1.0= 44.42mls
Design Wind Pressure at 10 m height (plJ
P, = 0.6 VZ2 [5.4]
plo = 0.6 (44.42)2= 1183.88 N/m2
30
/’
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]!
,...,, . &f~
SP 64 (S & T) :2001
Wind Load Calculation
F = (cPe - CPi)A.p~ [6.2.1, Tables 4 and5]
Internal Pressure Coetllcients (CPi)
Since openings are given as 5-20 percent of the wall areas, the value of C~i = M1.5[623 and 6.2.3.2]
The sign will depend on the direction of flow of air relative to the side of openings.
External Pressure Coefficients on Roof (CJ
The CPefor various sectors of the roof excluding the overhangs, are given in Fig. 16(a) and Table 7 as per[Table 5]. For Ww= 3.5/10 = 0.35, which is <0.5.
Table 7 Overall External Pressure Coeffkient
WindIncidencePressureChsfftdentPortionof Roof
@ w
E -0.9 -0.8F -0.9 -0.4
G -0.4 -0.8
H -0.4 -0.4
Pressure Coefficients for Overhang [6.2.2.71
External pressure coefficients are recommended as equivalent to nearby local pressure coefficients
[
-1.4, corner of width y = 1.5 mThus CPe=
–1 .2, other than comer 1However, for wind ward side of overhang the equivalent CPi(underside of roof surface) is +1.25 while onlee-ward side of overhang the CPiis equal to the Cw of adjoining wall.
Design Pressure Coeffkients on Roof (External)
CP(net) = CP – C@, are calculated for cladding (local pressure coeftlcients are given separately). The CPcalculations are summarized as follows:
If CPi= +0.5;
When 0 = OO;
CP(for region E and F)= -0.9- (0.5)= -1.4
CP(for region G and H)= -0.4- (0.5)= -0.9
When 9 = 90°;
CP(for E and G) = -0.8- (0.5)= -1.3
CP(for F and H) = -0.4- (0.5)= -0.9
If CPi= -0.5;
When 9 = OO;
CP(for E and F)= –0.9 – (-0.5) = -0.4
CP(for G and H)= 4,4- (-0.5)= +0.1
When f3= 90°;
CP(for E and G)= -0.8 – (-0.5) = -0.3
CP(for F and H) = -0.4- (-0.5)= +0.1
The CPvalues for various sectors of roof are given in Fig. 16(b).
31
..
SP 64 (S & T) :2001
The wind may come from either direction and opening may be on any side, hence the highest CPvalue for allzones of cladding is considered as –1.4 outward and+ 0.1 inward. Design will have to be checked for both thesepressures.
Design Pressure Coefficients on Roof (Local)
CP,local is negative everywhere. Depending on wind direction C~ican be +ve or –ve. Positive value of C~iwillincrease CP(net),which is to be considered for calculation of wind loads. The values of local CPare summarizedin Fig. 16(c).
Design Pressure Coeftlcients on Walls [6.2.2.1 and Table 4]
In this case h/w= 0.35 and VW= 1.8
— For these values of h/w and l/w [Table 4], the value of CP as per Codal provisions are given inFig. 16(d).
Value of CPiis the same as before = t 0.5
The CPcalculations are summarized below:
If CPi= +0.5;
When 6 = OO;
CP(for Wall A) = 0.7 – (0.5) = 0.2
CP(for Wall B) = –0.25 - (0.5) = -0.75
CP(for Wall C) = –0.6 – (0.5) = –1.1
CP(for Wall D) = -0.6- (0.5)= -1.1
When 0 = 90°;
CP(for Wall A) = -0.5- (0.5) = -1.0
CP(for Wall B) = -0.5 – (0.5) = –1.0
CP(for Wall C) = 0.7- (0.5) = +0.2
CP(for Wall D) = -0.1 – (0.5) = -0.6
If Cpi = ‘0.5;
When 8 = O“;
CP(for Wall A) = 0.7- (-0.5) = 1.2
CP(for Wall B) = –0.25 - (-0.5) = +0.25
CP(for Wall C) = –0.6 – (–0.5) = -0.1
CP(for Wall D) = –0.6 – (4.5) = -0.1
When O= 90°;
CP(for Wall A) = –0.5 – (-0.5) = 0.0CP(for Wall B) = -0.5 – (–0.5) = 0.0
CP(for Wall C) = +0.7 – (-0.5) = +1,2CP(for Wall D) = -0.1- (-0.5)= +0.4
The above CPvalues for walls are based on wind directions of (3= 0° and 0 = 90°. But with change of angle ofattack of wind, position of A–B and C–D get interchanged. Hence, maximum CPvalues for walls A or B = +1.2(and -1.0) and for walls C or D = +1.2 (and -1.1) are to be considered for the design of walls.
The comers of walls experience more force for which local <P is specified as under CP(Comer) = -1.0 -(+0.5)= –1.5 or -0.5. Hence, the four comers of the building for width of 0.25w that is 2.50 m are to be designed forCP=-1.5
32
/
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SP 64 (S & T) :2001
Computation of Frictional Drag [6.3 and 6.3.1]
For rectangular clad buildings when &h or &b >4 and h < b frictional drag force F’ (acting in the direction ofwind, for (3=90° and 270°) is given by
F’= c; (d - 4h) b.pd + c; (d-4h) 2 h.pd(drag on roof) (drag on walls)
Here, d=l=18m; b=w=lOm; h=3.5 m
C;= 0.01 (since external surface of walls is smooth)
Frictional Drag on Roof
= 0.01 (18 – 4 X 3.5)10 X 1 183.88= 473.55 N
Frictional Drag on Walls
=0.01 (18–4x3.5) 2x3.5 X 1 183.88= 331.50N
Total Frictional Drag = 805.05 N
The frictional drag on various structural components is to be distributed such that the first l/3rd windwardportion (6 m) should be able to resist half of frictional drag alongwith other loads and rest 2/3rd lee-wardportion (12 m) should be able to resist other half of frictional drag, on roof as well as on walls separately.
Final design pressure coefficient for cladding, fmtures etc, are given in Fig. 16 (e-g) and force calculationsare as follows:
Forces for regions other than local regions (for cladding and fixtures) considering higher value of CPfromFig. 16(e). Negative values mean that the roof or wall is being pushed outwards or upwards.
Force per unit area for roof when CP= –1.4
= -1.4x 1.0x 1183.88 = –1 657.43 N/m2
Force per unit area for longitudinal walls when CP= + 1.2
= 1.2x 1.0x 1 183.88= 1420.66 N/m2
Force per unit area for cross walls when CP= + 1.2= 1420.66 N/m2
Forces for local regions (for cladding and fixtures) are given in Fig. 16(f).
Forces for design of main frames etc.
For the design of main frames various load cases are given in Fig. 16(g). The values given inside the frame areto be used for calculating lateral force transferred from the side walls C or D. The frictional drag is to be takenappropriately alongwith these forces.
COMMENTS
(i) Each component located in a region is to be designed for the forces acting on it.(ii) The fasteners on cladding may be concentrated in the region of high wind pressure as shown in
Fig. 16(h).
33
SP 64 (S & T) :2001
1.5m
I
;
WIND INCIDENCE :ANGLE 0° I
I
;
;
l“5m
t-1’4
-OVER HANG
PLAN
16(a) ~~ for Roof
PLAN P1.NLocAL Cp c, ON OVERMAMG
16(c) ~ (Local) for RoofPLAN PLAN
-1.L -04
E,F G,H G,H E,FA
ELEVATION ELEVATIONMax. CP(8=O”) Max Cp(e=1800)
(cP;;)+ 0.5) (Cpi. t03)
(b]
16(b) ~ for Roof
FIG. 16 A RECTANGULARCLADBUILDINGWITHPITCHEDROOF(Continued)
,,,.
34
LOCAL W!NO(O.9(F’)cpe.-lo
C(-06) c +0.7) -1.0
03
Cp..f.o{
125 m
A (+07) B(-025)
WINOI@m
2 ~(0=0’)
1
&z m
& ;:* D(-O.l)
PLANPLAN SIDE
ELEVATION
FRONT ELEvATION
16(d) ~~ for Walls
WALL
SP 64 (S & T) :2001
WALL
m1
t rI I
1 I
I;I Uo OF :I1 :II -1.4,40.1 -1.L,40.J :I I
I: (- 165 7.43] :II I1 II I
t I
I I
c
tt
-f~+1.2,-1$1
L*(+1420.66)
IWALL
I
WALL
16 (e) ~ (Design) for Cladding
125-f1
L
~ W*FIGURESIN BRACKET INDICATE THE DESIGN
PRESSURE IN NII112
16 (f) ~ (Design-Local) for Cladding
FIG. 16 A REmANGULARCLADBUILDINGWITHPITCHEDROOF(Continued)
35
SP 64 (S & T) :2001
CASE(i 18=@, Cpi= +05 CASE (Ii) 0=0’, Cpi = -05
.CASE [“iii) e=gr)~c pi= +().5 CASE (IV) 9-900, Cpi=-o 5
:l:ls J $118 39)+0.1
(-1183.00)- F H - (-1163439)
— mF H-1.0 ‘ - -1.0 00 00
cASE(V) f3=90°, Cpi=+tJ.5 CASE(vi) e= d’, Cpi= -0.5
CASE (vii) AT ENDS CASE ( Viii) Af ErWSe=oe, cpi=+o.s e s9cf’, cpi = -05
FIGURES IN BRACKET INDICATE THE (.1!93-.:,:) -E G
DESIGN P~ESSURE IN N/m2--- f-1183.8B)
- -1.0-0.6, +0.2-
G710-33~236 .78).
CASE ( ix) AT EtIOSf3=900, Cpi= +05
16(g) ~ and Design Forces for Various Load Cases
FIG. 16 A REaANGULARCLADBUILDINGWITHPITCHEDROOF(Connnued)
,“
36
.SP 64 (S & T) :2001
1
—+++++*+ +++++++++ +++++++++ +++++++++ +++++++++ + +
+++++-++ ++{
+++++++ + +
+++++++ + +
+++++++ ++
+ + + +
+ + + +++++
+ +++
+ + ++
+ + + +
+ ++ +
+ + + +
+ + + +
—
?
—
16(h) Non-Uniform Distribution of Fasteners for Local Regions
FIG. 16 A RECTANGULARCLADBUILDINGWITHPITCHEDROOF
Example 2. A rectangular clad building with monoslope roof with overhangs (Fig. 17), [Table 6].
Given:
iio
1- -. -z L..-\_—-. — —.-— .-
FIG. 17 A RECTANGULARCLADBUILDINGWITHMONOSLOPEROOF
Physical Parameters:Height (h) : 4.0 mWidth (w) : 8.0 mLength (1) : 16.0 mRoof angle (~) : 20°
. ...— .
,—.
!.!
f’k +
!’
.’
37
SP 64 (S & T) :2001
Overhang : 0.5 mOpenings : Longer side of the building is open
External surface of wall : SmoothGround is almost flatLife 25 yearsReturn period 25 years
Wind Data
Wind zone: 2 (Basic wind speed 39 m/s)Terrain category :2 [5.3.2.1]Class of structure: A [5.3.2.2]
Design Wind Speed (VJ
V,= VWkl.kz.kJ
V~ = 39 mh [5.2, Appendix A]kl = 0.92 [5.3.1]k2 = 1.00 [5.3.2.2, Table 2]k~ = 1.00 [5.3.3.1]Hence, VIO= 39 x 0.92 x 1.0x 1.0= 35.88 nds
Design Wind Pressure (pZ)
pZ = 0.6 VZ2
PIO=0.6x (35.88)2= 772.42 N/m2
Wind Load Calculation
F= (CP – CPi)A.p~ (6.2.1, Table 6]
Internal Pressure Coefficients (CPJ[6.2.3.2 and Fig. 3]
Since one side of the building is open, and B/L or W/L <1, C~iare as follows from [Fig. 3].
[1+0.8 for (3= 0°-0.175 for e =45”
C~i = +.5 for e = 90”-0.5 for e = 135”-0.4 for e = 180°
Values of Etfor 45° and 135° are obtained by interpolation.
External Pressure Hlcienta on Roof (Cpe)
The Cw for various sectors of the roof and different wind direction angles [6.2.2.3 and Table 6] are given in Fig.17(a) and Table 8.
Table 8 External Pressure Coefficient on Roof (Cw)Overall CP for a = 20°
0 H L
o“ -0.8 -0.5
45° -1.0 -0.6
90° -0.9 for4 mfromwindward -0.9 for4 m fromwindward-0.5 for theremainingportion -0.5 for theremainingportion
135° -0.5 -1.0
18(T -0.2 -1.0
Pressure Coefficients for Overhang [6.2.2.7]
l“’
Pressure coefilcients on top side of overhang (equivalent C~ are recommended as nearby local externalcoefficients. However, for windward side of overhang the equivalent CPi(under side of roof surface) are
38I
I ..
SP 64 (S & T) :2001
recommended as 0.75 (for overhanging slope upward) and 1.25 (for overhanging slope downward) [6.2.2.7].On lee-ward side of overhang, the CPiisequal to the Cwof adjoining wall. The net maximum pressure coefficientsare given below and shown in Fig. 17(b).
Adjacent to Regions
I HI I He I L I LI I
I -1.8-0.75=-2.55 I -2.0-1.25=-3.25 \ -2.0-1.25=-3.25 I -1.8- 1.25=-3.05 I
It is to be noted that these values are the maximum coeficientsfor which the roof in thatportion shall be designed,regardless of the direction of wind. It is assumed that the under pressure on the side overhangs He and Le will
experience an upward force of 1.25, as also the overhang beyond L1 and ~, for some wind direction (for windfrom 180° in the latter case). Also that regions HzandLzare absent in the present example. Since the extent ofregion (Hl + .HJ is equal to W, the width of the structure is W = lJ2.
Design Pressure Coefficients on Roof
CP(net)= Cp. – Cpi,k calculated for different direction of wind as given below :
For 0 = OO;CP(region H) = –0.8 – (0.8) = –1.6CP (region J!,)= -0.5- (0.8)= -1.3
For 0 = 45°;CP(H)= -1.0- (0.8)= –1.8CP(L)= -0.6 – (0.8) = –1.4
For Cl= 90°;CP(windward 4 m) = -0.9- (-0.5) = –0.4CP(rest of the portion)= -0.5 – (-0.5) = 0.0
Fore = 135°;Cp (H)= -0.5 – (–0.4) = -0.1Cp(L)= -1.0 – (-0.4) = -0.6
For(3 = 180°;Cp(H)= -0.2- (–0.4) = +0.2Cp(L)= -1.0- (-0.4)= -0.6
It is clear from the above calculations that CPfor region H is –1.8 @ = 45°) and +0.2@= 180°) and for regionL is -1.4 @ = 450).
Local Design Pressure Coefflcienta on Roof
The net pressure coefficients on the edges of the roof and overhangs are shown in Fig. 17(b),
Pressure Coeftlcients on Walls [6.2.2.1 andTable 4]
Since one of the walls is open, some care has to be exercised in assessing the internal and external pressure onwalls. The internal pressures are to be taken from [Fig. 3], for the case B/L< 1.0. The external pressures for thesolid walls as well as the pressures in the local region at the comers are to be taken from the appropriate casesof [Table 4]. The external pressures for the case hlw = 1/2 and Uw= 2.0 are shown in Fig. 17(c). Taking theinternal pressure from [Fig. 3], the net pressure on the wall is given in Fig. 17(d). Design must be checked forall the wind loadings conditions (positive and negative) and for each wind direction. CPfor comers of wall aregiven in Fig. 17(e).
/
Since &h or db is not greater than 4, the building will not experience frictional drag.
39
,
SP 64 (S & T) :2001
WHEN 0#90°( REFER TABLE2-1)
– 0.9
ie=90° &b21n
w 1-.WHEN 13=90° LOCAL Cpe
17 (a) ~ for Rcmf
“m :T-3”2’l-- 4.05
-2.6
2,8lzzzzzK-\\ ;......: ..!.
. .
:. .
. .
;.. .
,.
. “.:
~
i-s---iv‘-- ~-2.6
-2.55
H
-3.05“.
-3.25 -3.25
L-=--lLOCAL CP ON ROOF Cp ON OVERHANG
17 (b) ~ (Local) for Roof
FIG. 17 A RECTANGULARCLADBUILDINGWITHMONOSLOPEROOF(Conlinued)
. .
,,,’
40
-0.6\\\\ c \\<
//+0.8
LOCALREOIONS(CLADDING)
~ (oA~N)B
[\D \\\y
-0.6
Cpe (e=d’)
-0.6
1(O$EN)
-0.25 B—(W:N)
D
-0.1
Cpe ( e =90°)
SP64(S&T):2001
-0.5
I +0.7
-!.0
-0.6 u
Cpe(Q=180°) LOCAL Cpe
17 (c) %* for Walls
FIG. 17 A RECTANGULARCLADBUILDINGWITHMONOSLOPEROOF(Continued)
,,..
41
SP 64 (S & T) :2001
-1.4
c
e
D-1.4
CP (e=o”)
0.2
c
B
o
-0.2
CP (e=lecf)
+1.2
.1-05 A B
D
+0.4
CP (e=9d’)
+1.1
0.0
17 (d) ~ for Walls for Various Angle of Attack of Wind
FIG. 17 A RECTANGULARCLADBUILDINGWITHMONOSLOPEROOF(Continued)
!,
,.,,, I
42
,+1.8
iI1I
+III1iI1iI
i-J
\+l.8
+1.8
*
===6°”’
D -0.6
I-.--—- *IIIIIII
-0.5 III1
~
I-------- 4
D
SP64(S&T):2001
17 (e) ~ on Comers of Vertical Walls
FIG. 17 A RECTANGULARCLADBUILDINGWITHMONOSLOPEROOF
Example 3. The rectangular clad building of Example 2 has been considered to be situated on a hill having aslope of 9.5° on one side and 19.(Y on the other with walls on all sides with opening less than 5 percent [5.3.3.1and Appendix C].
Given:
The approximate features of topography are given in Fig. 18.
The building is located at 40 m from mean ground level.
The slopes on either side of crest are greater than 3°, hence it is considered as hill.
In such cases, the total methodology of solution remains the same except that the value of ks is to becalculated as per [Appendix C].
$
>.
.,,
43
/
SP 64 (S & T) :2001
WIND~’””’”’!
Z=loo
,EL. 200
18(a) Wind Direction from Left to Right
‘7”’”’” z.—.—.
2=100
EL. 200
w’””’”~
EL.200
(b) Wind Direction from Right to Left
FIG. 18 A BUILDINGWITHMONOSLOPEROOFONA HILLSLOPE(Continued)
When Wind Blows Left to R@t — Fig. 18(a)
Upwind slo@. e = 9.5°Downwind slope 0 = 19.0°Effective horizontal length, L,= L = 600 mDistance of building (X) from crest = – (L – Z-Zcot El)
= - (600 -40 cot 9.5°) = -360 mH/Le = 0.067 and X/Le = –0.60
(Since the building is situated on upwind slopti, negative sign is used)
k3=l+c,Fore = 9.5° (3° <6< 170), C= 1.2 (100/600)= 0.2
s = 0.20 [Fig. 15]
Hence k3 = 1 + 0.2x 0.20= 1.04
When Wind Blows from Right to Left — Fig. 18(b)
Upwind slope (1= 19°Downwind slope f3= 9.5”Effective horizontal length L,= z/0,3 = 100/0.3= 333.33 mDistance of building (X) from crest= +360 mH/L, = 0.12 and X/Le = +1.08
Iq=l+c,For 9 = 19°>17°, C = 0.36, s = 0.35 [Fig. 15]Hence k3 = 1 + 0.36x 0.35= 1.126
. .
NOTE—Fordesign,greaterofhe two vafues of kqis considered.
44
SP64(S&T):2001
Thus,V10=39 X0.92 X 1.OX l.126=40.401n/sp10= 0.6 x (40.40)2= 979.34 N/m*
The rest of the procedure is similar to those used for Example 2 except that the pressure on all the walls shallbe taken from [Table 4] only (not on three walls only) and the internal pressure shall be taken as recommendedin [6.2.3], one observes that the internal pressure coefficient can be either +0.2 or -0.2 and the value whichproduces greater distress must be considered for design.
External Pressure Coefficient on Roof
These values shall be taken as given in Fig. 18(c) together with the Table 9 given in it.
NOTES
1 Wall pressures indicated when wind is coming at (1= 180”,are the reverse of those given at 61= W, that is, +0.7 on B, -0.25 on A,
-0.6 on C and D, –1.0 on the near edge F.
2 Wall pressures indicated when wind is coming at El= 27LTare the reverse of those given at (1= 90°, that is, +0.7 on D, –O.1 on C,-0.5 on A and B and -1.0 on edge H.
Table 9 External Pressure Coefficient on Roof (Cw)
EdgesOverhange H L
Ht He L L1 El Ez E3 E4 E5 E6
& -0.8 -0.5 -1.8 -2.0 -2.0 -1.8 -1.8 -2.0 -2.0 -2.0 -2.0 -1.8
45” -1.0 -0.6 -1.8 -2.0 -2.0 -1.8 -1.8 -2.0 -2.0 -2.0 -2.0 –1.8
90° -0.9 -0.9 -1.8 -2.0 -2.0 -1.8 -1.8 -2.0 -2.0 -2.0 -2.0 -1.8(4 m Windward)
90” -0.5 -0.5 -1.8 -2.0 -2.0 -1.8 -1.8 -2.0 -2.0 -2.0 -2,0 -1.8(4 m to 16m)
135” -0.5 -0.5 -1.8 -2.0 -2.0 -1.8 -1.8 -2.0 -2.0 -2.0 -2.0 -1.8
180° -0.2 -1.0 -1.8 -2.0 -2.0 -1.8 -1.8 -2.0 -2,0 -2.0 -2.0 –1.8
Fig. 18(d) gives the external pressures on vertical walls.
Design Pressure Coefficient
From the values of the external and internal pressures coefficients, the following design pressures coefficientsare worked out:
Walls
Face A {(+0.7 - (-0.2)} = +0.9 and {-0.2 - (0.25)}= -0.45Face B {+0.7 – (-0.2)] = +0.9 and -0.45Face C+ 0.9 and {-0.6 – (0.2)) = -0.8Face D + 0.9 and -0.8Edges E’, f? and F, F, –1.2,Edges G, G’ and H, H’, –1.2.
Roof (excluding overhangs)
H, M, L, L’ = -1.2HI, HI’, Ll, L1’ = –2.0He, He’, L=,Let = –2.2
Overhang
\I
,1
Some care is required in assessing the under pressure on the comer pressure E2 and Es. They are taken asequivalent to ‘sloping downwards’ [6.2.2.7] for wind coming from Et= 90°, to be conservative. Portion El,definitely slopes upwards for wind at (3= 0° and E6 slopes downwards for wind at e = 180°. These lead to thefollowing design pressures forE, and Eb = (- 1.8- 0.75) = -2.55, Ez to Es = -2.0-1.25 = -3.25. The shear onthe foundation is observed to be maximum at 0 = W.
45
I
SP 64 (S & T) :2001
E&
b
He : Le
t
.7 ---- -_--T-
;I1 .:
I 1 :;,I
: I .; :HII H
I L ~Lj ;;:I I ..:
! : :?I .. .1 .. .1’
t-- “
I .....~.
8I
,: !{I ; H’ ‘L’l
II II II
:
.1---- ‘-_-- l-_
r
I
18 (c) ~e on Roof, its Edges and Over Hanging Portion All Round
-0,6
rwl
E F
y ---c ?
i I! 1
-0.7 1‘-OVER HANG1 G’ -0.5 H’I
1A 30,2 ~uuu__-J. -__-- ___ L--
1 B1
I I
-&6
— 0.25
II
m
1.?E IF
I
9 =0°( FROM HIGHER SIDE)
‘0’7--0”.---- ----- ----- ----
e =90°
1’
/!”
18 (d) Pressure on Vertical Walls
FIG. 18 A BUILDINGWITHMONOSLOPEROOFONA HILLSLOPE
46
I,j
I
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SP 64 (S & T) :2001
Example 4. A multispan saw tooth shaped roof building (Fig. 19) [Table 17].
Given:
Physical parameters:Height (h) : 6.0 mWidth (w’) : 12.0 mLength (w) : 60.0 mNo. of Units : 4Roof angle ((x) : As shown in Fig. 19
Openings on sides : Maximum 5-20 percent of wall areaExternal surface of walls : Rough (Corrugated)
1 w’= 12; 1 W’= 12m-1 =1 i
6
b
(1
ROOF PLAN
*a b~ dm n x
;+
7s.7°7.3°~ +“,. ,
6.0 m
:‘-L
1 1 1 1 Lr
EO. EO.1 Ea. EQ.
1 ‘1 1 v
,..,!
ELEVATION
FIG. 19 BUILDINGWITHMULTISPANSAWTGGTHSPEEDRGGF(Continued)
47
/
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sP64(s&T):2001...
Wind Data
Wind zone: 2 (Basic wind speed= 39 M/s)TerrainCategory: 2 [532.1]Class of structure: C [53.2.2]Topography:Almost flat
Design Wind Speed (VJ
v, = v~.k~.k*.k3v~ = 39 rn/seck, = 1.00k2 = 0.93k3 = 1.00V, = 39x1x0.93x1 = 36.27 m/s
Design Wmd Pressure (@
2 06 (36.27)2= 789.31 N/m*pZ= 0.6 X (Vz) = . ,
Wind Load Calculation
Internal Pressure Coefficient [6.23 and 6.232]C@= M.5
External Pressure Coeflcient [6.22.6 and Table 17]
Region a b c d m n x z Local
Cw for(3=0° +0.6 -0.7 -0.7 -0.4 4.3 -0.2 -0.1 -0.3
1-2.0 and -1.5
e = 180° -0.5 -0.3 -0.3 -0.3 -0.4 4.6 -0.6 4.1
Cw foro = 90° -0.8 -0.6 -0.20 = 270° -0.2 -0.6 -0.8
Design Pressure Coe@cients on Roof
Cp(net)= Cw – C@
h=6m, h1=h2=h=6m, h~=48mWidth on either side of ridge = 0.1 w’= 1.2 m
Cp (net) for roof
Region a b c dm n x z Local
CPfor (3= 0° and Cpi = +0.5 +0.1 -1.2 –1.2 -0.9 -0.8 -0.7 -0.6 -0.8 —CPfor 9 = 0° and Cpi = -0,5 +1.1 -0.2 -0.2 +0. 1 +0.2 +0.3 +0.4 +0.2 —Cp for O= 180° MCICpi=+0.5 -1.0 --0.8 -0.8 -0.8 -0.9 -1.1 -1.1 -0.6 —CPfor (3= 180° and Cpi=-0.5 0.0 +0.2 +0.2 +0.2 +0.1 -0.1 -0.1 +0.4 —
Design CP (net) for (3= 0° and 180”
Region a b c dmn x z Local
{
,,
/
Design
I+1.1 +0.2 +0.2 +0.2 +0.2 +0.3 +0.4 +0.4 - -2.5
Pressuret
andCoefficient -1.0 -1.2 -1.2 -0.9 -0.9 -1.1 -1.1 -0.8 -2.0
48
SP64(S&T):2001
These coefficients have been marked in Fig. 19 (a) and (b).
CP(net) fore= 90°
Region hl hz hs
when,C~i= +0.5 ‘1.3 -1.1 -0.7C~i= ‘0.5 +0.3 -0.1 -0.3
These coefficients have been indicated in Fig. 19(c).
Final design CPare:
Region a b c d m n x z
Cp +1.1 –1.2 –1.2 –0.9 -0.9 –1.1 –1.1 +“.8
The design coefficients for roof cladding are given in Fig. 19(d).
Design Pressure Coefficient for Walls [6.2.2.1 and Table 4]
As per the definition given in [Table 4].w=48m, l=60m, h=6mh/w = 0.125 and VW= 1.25
For Wws 0.5 and 1< l/w S 1.5, values of CPare
Region A B c D Local
CP.for9=0° +0.7 -0.2 -0.5 -0.5 -0.8C~l=+0.5 4.5 -0.5 -0.5 -0.5 -0.5C~i=–0.5 +0.5 +0.5 +0.5 +0.5 +0.5
+0.2CPfor 0=0° +1 z
-0.7 -1.0 -1.0 -1.3+0.3 +0.0 +0.0 +.3
e = 90° -0.5 -0.5 +0.7 -0.2 -0.8C~i‘4.5 +0.5 +0.5 +0.5 +0.5 +0.5C~i=+0.5 +.5 -0.5 -0.5 4.5 -0.5
Cp, e = 90° +0.0 +0.0 +1.2 +0.3 -0.3–1.0 -1.0 +0.2 -0.7 -1.3
Considering the variability of the wind direction, the C (design) for all sides is +1.2 and –1.0 and CP(Local)–1.3 for comer strips on all four corners of 0.25w = 1~ m width-wise and 0.251 = 15 m length-wise and areindicated in Fig. 19(e). Also extreme values have to be recommended for the range of internal pressures.
Calculation of Frictional Force [6.3.1 and Table 17]
When wind blows parallel to longer wall, that is, 8 = 90° and 270°, then
h=6m, b=4, w’=48m, d=l=60moYh=lOandh<b
The total frictional drag force (on both the vertical walls and roofs) is given byF = C/ (d – 4h) b.p~ + C: (d – 4h) 2h.p~
drag on roof drag on wall
Here C;= 0.02 for corrugated sheets as external surface with corrugations across the wind direction,F'= 0.02(60-4x 6)x48 x789.3 l+ 0.02(60-4x 6)x2x 6x789.3l
= 27 278.55+6 819.64=34 098.19
h!” “!’, ‘#,..,, /,
J
. .———
-.
v ——
I
$,
.,
49
/
SP 64 (S & T) :2001
The frictional drag force on various structural elements is distributed such that 50 percent of this frictional dragis resisted by windward l/3rd length (20 m) alongwith other loads and balance 50 percent force by rest 2/3rdlength (40 m) in each direction O= 90° and 270°.
Frictional Drag Force for Roof
Frictional drag force =27 278.55 N, half of this, that k27 278.55/2= 13639.28 N is to be resisted by the ama = 20.00 x 48 m2Force per unit area = 13 639.28/(20.00x 48)= 14.21 N/m2
13639.28 N is to be resisted by another 2/3rd length that is, 40.00 mForce per unit area = 13 639.28/(40.00x 48)= 7.10 N/m2
The frictional force distribution is indicated in Fig. 19(f).
Frictional Drag Force for Wall
Frictional drag = 6819.64 N3409.82 N is to be resisted by 20.00 m lengthForce per unit length = 3 409.82/20.00= 170.50 N/m
Other half 3409.82 N is to be resisted by 40.00 m lengthForce per unit length = 3 409.82/40.00= 85.25 N/m
When wind blows parallel to width ((3=0° and 1800), then
h=6m, b=l=60m, d=4w’=48m, a7h= 8andh<b
~ = 0.02 (48-4X6)X60X 789.31+ 0.02 (48 -4X6)X2X6X 789.31= 22.732+ 4.546= 27.278
This force is distributed such that half of this frictional drag is to be resisted by windward l/3rd length (16 m)and balance by rest 2/3rd length (32 m) in each direction t3= 0° and 180°.
For Roof
11366.0 N is to be resisted by the l/3rd length that is, 48/3 = 16 mForce per unit area = 11 366.00/(16.00x 60)= 11.84 N/m211366.00 N is to be resisted by the 2/3rd length that is, 48 x 2/3 = 32 mForce per unit area = 11 366.00/(32.00x 60)= 5.92 N/m2
The frictional force is indicated in Fig. 19(g)
For Wall
2273.0 N is to be resisted by the length 20.00 mForce per unit length = 2 273.0/20.00= 113.65 N/m
2273.0 N is to be resisted by the length 40.00 mForce per unit length = 2 273.0/40.00= 56.83 N/m
Calculation of Force
F = CP.p~ per unit area wherep~=789.31 N/m2
I c“ I Force, N/m2 I CtI I Force, N/m2 I
+1.1 868.24 -1.3 -1026.10
-1.2 -947.17 -0.7 -552.52
-0.9 -710.38 -2.5 -1973.28
-1.1 -868.24 -2.0 -1578.62
I
,,
I 1 1
-0.8 -631.45 +1.2 +947.17
50
I
SP64
(S&
T)
:2001
I4
\\
\W
Fx’i’’’’’’’’’’’’””
M
r-’iI
0
I0NI
o
01+
J~
E0N.~
Eo&d.
foN
r-
E0&9
,.,’,
51
I
---
ulIQ
—.,
k 12.Om 1 lz.orn L 12.Om i 12.Om‘1 ? . J
19 (c) ~ (Overall) e = 90° Roof Plan
60.Om
:
-1.1
I&l
-0.7
— -7.5 —-2,0 1 ~ -2.5
1 lzlom 1 12.Om 1 12.Om 1 12.Om-1 -1 1 7 4’
19 (d) ~ Design for Roof Cladding
Neow
13G.19 BUILDINGWITHMULTISPANSAWTOOTHSPEEDROOF(Continued)
..
__—.——
,––-–. ———— .’- “q!m=%R-i-”’; ~
.~.>.. -..+’
“\
-A&&a.&.—a– i—. .. .. .%-;.,:.-s=s..+ !.
. .. _ ,.._ ._, - ,..
I
SP 64 (S & T) :2001
I%E-,:--l;1.3
h=6.Om
I
J 12.0 136.Om LT 1
Cp ( LOCAL) WALLS 6=0°
-,Cp WALLS e = 0°
15.0 [ 45.0 m L-7” 1
1
Cp (LOCAL) WALLS e=W”
d-I.oTB
+
h:60-1.0 I
-~4 -.68.0m
1 1
Cp wALLS e=90°
19 (e) ~ for Walls
FIG. 19 BUILDINGWITHMULTISPANSAWTooTH SPEEDROOF(Continued)
1
/
!,,.
,.
,
53
I I
I II
---
~11+.zl N/mz . ~lls8k N/mz m~5”92 m
m
19 (f) Frictional Force for Cl= 90°
3
4 16.Om ~ 32.Om 11 1
19 (g) Frictional Force for 0 = 0°
FIG. 19 BUILDINGWITHMULTISPANSAWTOOTHSPEED ROOF
●
.
.. .,
___ .—— —————.——..
..
,,,
SP 64 (S & T) :2001
Example 5A rectangular clad building with multispan pitched roof (Fig. 20), [Table 16].
ROOF PLAN
mm
a bc d~ I!g 1
If~-th=6.Om
aIs: a! “ --L
L Zoom ~Om 20.0 ● 20.Om
1 1 ‘1 7 7ELEVATION
FIG. 20 A BUILDINGwrn+ MUL’MSPAN~CHED RGGF(Continued)
Given:
Physical Parameters:Height (h) : 6.0 mWidth (w’) : 20.0 mLength (w) : 100.0 mNo. of Units : 4Roof angle (00 : 12°
Openings on sides: Less than 20 percent of wall areaExternal surface of walls: Rough (Corrugated)Number of gable frames in length of 100 m = 21All spans being equal.
55
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i
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sP64(s&T):2001
Wind Data
Wind zone: 2 (Basic wind speed= 39 rids)Terraincategory: 2 [5.321]Class of structure: C [5.3.2.2]Topography:Almost flat
Design Wind Speed (V)
Vz = vb.kl.k.#3v~ = 39 In/skl = 1.00k2 = 0.93k~ = 1.00V, = 39xlx0.93x l=36.27m/s
Design Wind Pressure (pZ)
Pz = 0.6X (VZ)2 = 0.6 (36.27)2 = 789.31 N/rn*
Wind Load Calculation
Internal Pressure Coefficient [6.2.3 and 6.23.2]
Since openings are less than 20 percent of wall area the vrdue of CPi= t 0.5
External Pressure Coefficient [6.2.2.6 and Table 16]
Since the values of pressure coefficients are given in [Table 16] for roof angles of 5°, 10°,20° ...... the valuesof CP for ct = 12° are obtained by interpolating between the values at a = 10° and tx= 20°.
Region a b c dmn x z LQcal
Cp for0=0° –1.02 -0.6 -0.4 -0.3 -0.3 -0.3. -0.3 -0.42
I–2.0 and –1.50 = 90” -0.42 -0.3 -0.3 -0.3 -0.3 -0.4 -0.6 -1.02
Region h, hz h~
CP, fore = 90° -0.8 -0.6 -0.20 = 270° -0.2 -0.6 -0.8
Design Pressure Coefficients on Roof
Cp(net)= Cw - CPih=6m, h1=hz=h=6m, h~=88mWidth on either side of ridge, y = 0.1 w’= 2.0 m
Region a b c d m n x z Local
CPfore= 0° and Cpi = +0.5 +1.52 –1.1 -0.9 -0.8 -0.8 -0.8 4.8 -0.92
CPfore= 0° and Cpi = –0.5 -0.52 -0.1 +0.1 +0.2 +0.2 +0.2 +0.2 +0.08
CPfor t3= 180°and Cpi = +0.5 -0.92 -0.8 -0.8 -0.8 -0.8 -0.9 -1.1 –1.52
CPfor 0 = 180° and Cpi= -0.5 +0.08 +0.2 +0.2 +0.2 +0.2 +0.1 -0.1 -0.52
Design ) –1.52 –1.1 -0.9 -0.8 -0.8 -0.9 –1.1 –1.52 ] –2.5Pressure
‘1 Iand
Coeff. +0.08 +0.2 +0.2 +0.2 +0.2 +0.2 +0.2 +0.08 -2.0
.,’
These coefficients have been marked in Fig. 20(a) and (b).
56
I
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! SP 64 (S & T) :2001
Region hl h2 h~
CP for El = 90° and 270°Cp= +0.5 –1.3 –1.1 +.7
1 CPfore =90° and 270°b Cp= -0.5 +0.3 –o. 1 -0.3
1 These coefficients have been indicated in Fig. 20(c).t
The design coefficients for roof cladding are given in Fig. 20(d) and (e).
Design Pressure Coej6cientfor Walls [6.2.2.1 and Table 4]
W= gem, 1= loom,” h=(jm, h/w= ().()75and//w= 1.25
For h/ws 0.5 and 1< l/w< 1.5, values of CPare
Region A B c D Local
CP, for0=0° +0.7 -0.2 -0.5 -0.5 -0.8Cpi= +0.5 4.5 4.5 4.5 4,5 4.5Cpi= ‘0.5 +0.5 +0.5 +0.5 +0.5 +0.5
Cp, 0=0” +0.2 -0.7 –1.0 –1.0 -1.3+1.2 +0.3 +0.0 -0.0 -0.3
CP, fore = 90° -0.5 -0.5 +0.7 -0.2 -0.8Cpi‘4.5 +0.5 +0.5 +0.5 +0.5 +0.5Cpi‘+0.5 4.5 ‘4.5 4.5 4.5 4.5
.
Cp,e = 90°+0.0 +0.0 +1.2 +0.3 -0.3–1.0 –1.0 +0.2 -0.7 –1.3
Considering the variability of the wind direction, the C (design) for all for sides is +1.2 and –1.0 and CP(Local)–1.3 for comer strips on all four comers of 0,25w = 26 m width-wise and 0.25 x 1 = 25 IP length-wise and areindicated in Fig. 20 (d) and (e).
Calculation of Frictional Force [6.3.1 andTable 17]
When wind blows parallel to length that is, O= 90° and 270°, then
h=6m, b=4w’=80m, d=l=lOOm, a7h=16.67>4
The frictional drag force is given by
F = C; (d - 4h) b.p~ + C: (d - 4h) 2h.p~drag on roof drag on wall
Here C;= 0.02 for corrugated sheets as external surface with corrugations across the wind direction ,
F = 0.02 (100 - 24)X80X 789.31+ 0.02 (100-4X6)X 12X 789.31 =95 980.09 +14 397.01 = 110377.1.
The frictional drag force on various structural elements is distributed such that 50 percent of this frictional dragis resist by windward l/3rd length (33.33 m) alongwith other loads and balance 50 percent force by rest 2/3rdlength (66.67m) in each direction (1= 90° and 270°.
For Roof
I
i
Frictional drag force =95 980.09 N95 980.09/2=47 990.05 N is resist by Area = 33.33 x 80 m2Force per unit area =47 990.05/(33.33 x 80)= 17.99 N/m247990.05 N is resist by another 2/3rd length that is, 66.67 mForce per unit area = 13 639.28/(66.67x 80)= 8.99 N/m2
57
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SP 64 (S & T) :2001
For Wall
Frictional drag = 14397.01 N7198.50 N is resist by 33.33 m lengththat is Force per unit length= 7 198.50/33.33 = 215.97 N/mAnother half that is, 7189.50 N is resist by 66.66 m lengththat is, Force per unit length = 7 198.50/66.66= 107.97 N/m
When wind blows parallel to width@= 0° and 1800), then
h=6m, b=l=lOOm, d=4w’=80m, ti= 13.33, &’b=0.8, ti>4
~ = 0.02 (80 -4 X6)X 100X 789.31+ 0.02 (80 -4 X6) X12X 789.31=88 402.72+ 10 608.33=99 011.05
This force is distributed such that half of this frictional drag is resist by windward l/3rd length (26.67 m) andbalance by rest 2/3rd length (53.33 m) in each direction Et= 0° and 180”.
For Roof
44201.36 N is resist by the l/3rd length that is, 80/3 = 26.67 mForce per unit area =44 201.36/(26.67 x 100)= 16.57 N/m*44201.36 N is resisted by the 2/3rd length that is, 80 x 2/3 = 53.33 mForce per unit area =44 201.36/(53.33 x 100)= 8.29 N/m*
For Wall
5304.2 N is resist by the length 26.67 mForce per unit length = 5 304.2/26.67= 198.88 N/m5304.2 N is resist by the length 53.33 mForce per unit length = 5 304.2153.33= 99.46 N/m
Calculation of Force
F= CP.p~ per unit area, where p~= 789.31 N/m*
Cp Force,N/m2 Cp Force, N/m2
+1.2 947.17 -1,3 -1,026.10
–1.52 -1199.75 -2.5 -1973.27
-1.1 -868.24 -2.0 -1578.62
-0.9 -710.38 -1.0 -789.31
-0.8 -631.45 +1.2 +947.17
-0.7 -552.52
I
I
58
I
Inq4
SP64(S&
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~’~’~=~
b1
..--—
.----
---—-
--------—
—-—
——
----
#t
\ \\\
-----------
---------—
------
-----—
Yo
~+
1’4
I-----------------—
------
----——
----1
\\la
#t88
\b*.
-m
‘-ii ‘3
---—----—
-------‘s
!
59
SP 64 (S & T) :2001
r-!.1 r -1.3
1 W~ zom 1 WA20H 1 ti=zom 1 W’-zom7 @l 1 7 1
20 (c) Roof Plan ~ (Overall) for El= 0°
6
6
W=loom
I-0.7
(-1!92.72)-152
/
jy””;? t! : t “;:”+ ::+“LON7WALLS10SWALL 26.67 1 53:33m -1.0, (-769.31 )
(7b7.172),+1.2 a ‘60.0 m
20 (d) Design Wind Presure for Walls and Roof
FIG.20 A BUILDINGWITHMULTISPANPITCHEDRGGF(Continued)
60
. .
,,”
/
2t5.97 NlmWE WALL-1.0
L Cp* +1.2 (947.17) e= 96
20 (e) ~ for Walls and Rmfs
FIG. 20 A 13UILDINGwrrH MULTISPANPITCHEDROOF
Example 6. A rectangular clad goabwn buiiliing with combined roof (Fig. 21) [Table 20].
Case 1 — If the Annexe is clad.Case 2 — $the Annexe is unclad (sides open)
Given:
Physical Parameters:Height of main building (hl) :Height of annexe building (hz):Width of main building (b~:Width of annexe building (bl) :Slope of roof of main building a :Slope of roof of annexe building et:Length of building (1):Permeability:
Wind Data
Wind zone :2 (basic wind speed= 39m/s)Terrain category :2Class of structure: AProbable design life of structure: 50 years
Design Wind Speed
v~ = 39 m/seekl = 1.00k2 = 1.00kj = 1.00Vz =39xlxlxl=39rrtls
Design Wind Pressure
pZ = 0.6(39)2 = 912.6 N/m2
Internal Pressure CoeftlcientsC“i= * ().7
8.0 m4.0 m4m3m20°18.4°18.0 m>22 percent of wall area
E;ternal Pressure Coeftlcients for Roofs [6.2.2.2 andTable 5]
Main Building
h/w=8/(4+3) =8/7=1.14, ct=20°andy =0.15 x7=l.05mNOTE — It is not clear from the Code, as to whether the lesser horizontat dimension as seen in the plan including the annexure or ofthe portion abnve the annexed should be taken for w. It is recommended that the lesser horizontrddimension at ground level be taken ifhl > (),2h2, If hl c 0.2 h2, w Shou]d k me lesser horizontal dimensiwr of the portion above the annexure.
61
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SP 64 (S & T) :2001
Region EF GH EG FH
wind angle wind anglee=o” 9=90°
~ = 2(3” -0.7 -0,5 -0.8 -0.6
Figure 21 (a) and (b) give the Cp values for main building including local values.
CP, on roof of annexe building and vertical common wall [6.2.2.10 and Table 20]
Here ctl <30°, bl < bz and hl/h2 = 8/4 = 2.0CPe= 0.4 hllhz – 0.6 = + 0.2
For portion a CW = +0.2 for direction 1Cw = –0.4 for direction 2
For portion b CF = +0.7 for direction 1Cpe= –0.4 for direction 2
For portions c and d [Table 5], as given above
Cp~for Walls [6.2.2.1 and Table 4]
h/w= hl/(bz+bl) =7/5 = 1.14,0.25w =0.25x7= 1,75 mat(3= O°and4.5 mate =90° andfi= 18/7=2.57
I
wALL ‘k r10.Om
L
,
7L
bt= 3.0 m bz - 6.Om1
ELEVATION
WALL°C’
IIiIIIIIIIIIII
[II
-
WALL ‘B’
,,’
PLANFIG. 21 PI’ITHEDROOF WITHA VERANDAH(Continued)
62
sP64(s&T):2001
Region A B CD Localwhen(3=0° +0.7 -0.3 -0.7 -0.7e = 90” -0.5 -0.5 } –1.1
+0.7 -0.1
CP(net) for portions c and d [Table5]
CP(net) = CW– Cpi and Cpi = M.7
For (3= 0°, Cp (net) for region EF = -0.7-0.7 = – 1.4CP (net) for region GH = – 0.5-0.7 = – 1.2Cp(net) local= –2.2 and -1.7
Foro = 900, Cp (net) for region EC = -0.8- 0.7= –1.5Cp(net) for region FH = -0.6- 0.7= -1.3Cp (net) local= –2.2 and –1.7 for width of 0.6 m
Cp(net) for portions a and b
For portion a,
when 9 = 0° Cp = +0.90 or Cp = -0.50when (1= 180° Cp=–l.lOr CP=+0.3
For portion b,
when (3 = 0° Cp= +1.4 or Cp= 0.0when O= 180° Cp=–l.lor Cp=+0.3
Cp(net) values for roof are shown in Fig. 21 (c) and (d)
Cp(net)for Walls
Region A B c D Local
When (3= 0°+0.0 -1.0 -1.4 -1.4+1.4 +0.40 +0.0 +0.0
–1 .8) 4.4
e = 90°–1.2 –1.2 +0.0 -0.00+0,2 +0.2 +1.4 +0.60
These values of C~net) on walls am shown in Fig. 21 (e)
It is to be noted that the local pressure of –1.8 must be apphcu uu UIC VICIUCUI wau puIuuu u tusu, us uu UIC IIUIU
and back walls, at (3=90° and 270°. In many structures, design will have to be checked for both +ve and -ve CP- I.S-l f- -1.0 ~y<h<045w
- y80.60m
Orn
,,,,
,..
(a) Cpe VALUE FOR ROOF(c,d ) (b) Cp. VALUE FOR ROOF (c)d)(e=o”) (e=90°)
FIG. 21 PrrcHEDRGGFWITHAVERANDAH(Continued)
63
SP 64 (S & T) :2001
~
T“’ 7:-- “;-?
especially when there are load carrying frames, since some members may go into buckling for one of the loading ‘,,“’
conditions. k- -—I --Design pressure coefficients are shown in Fig. 21 (g).
If annexe portion is open except for the overhanging portion, it is treated as overhang [6.2.2.7] having Cpi= +1.25.
Hence CP(net) = 1.45
The portion of wall b below the overhang is to be designed for a pressure coefficient Cw of 0.9 as in [Table 21,portion Hl for wind from left to right and CPi = f 0.7 inside pwsure and Cw = -0.4 and Cpi= + 0.7 for windfrom right to left. Although net pressures are generally given at 0 = 0° and 90°, the design shall be checked atO= 180° and 270°, unless the structure is built symmetrically in both direction.
NOTE — Thatthe ‘Local’presswesat the comers have afsoto be assumedto be carried over to the portionsB and C and D of thevertical walls of the main building, in addition to the local pressures at the comers of the boundary of the entire building.
-2.27 f- -1.7
l--=--l(C 1 Cp DESIGN FOR Mw(c,d) (d)
INCLUDING LOCAL(e-o”,90”)
m
1=3.0 m 4.Omi
Cp DESIGN FOR PORTION ‘a’ OFROOF { 8= O“)
21 (a – d) Main Building Including Local Values
(-1.0 AND+W -1.0 (+1.4e=leoe) +0.4
+
.0
pa
Im.m
0=0”
‘w-
1 e.lelti
175 m 1.75m “ 1.75m
~~~] AT 0-0”1 1-148 \I&,~]AT e=moo
AT 9-M’)
,’”
21 (e) Net Pressure Coefficients on Vertical Wallsat t3= 0° and 180°
FIG.21 PITCHEDROOFWITHAVERANDAH(Continued)
64
I
-1”8
-1.2 ~ +1.4 ,+0.4
b-
A-098
SP 64 (S & T) :2001
21 (f) Net Pressure Coefficient on Vertical Wallsate = 90”
RG. 21 PITCHEDRGGFWITHA VERANDAH(Conlinued)
/
i
,.
,,,’
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SP 64 (S & T) :2001
3.0 m L b.o m k5 1 1
~ -1-8
L—.—
a
T-t.—————I
—.-t
Jc d
4.om
L -1,8 FOR 0=$0014 FOR 9=0”
e=
- ...—.4 .,,
.
-—
i!
-1.8
--1.8
1 3.Om 1 4.OI!I L-3 . ~
21(g) Pressure Coefficients on Vertical Walls at (3= 0° and 90°
FIG. 21 PITCHEDRDDFWITHA VERANDAH
Example 7. A rectangular clad building with roof having sky light (Fig. 22), [Table 20].
Given:Physical Parameters:Height (hl): 6mTotal height of structure (hz) : 6.8 mWidth (bl): 5mWidth of sky light portion (bz): 2 mTotal width of structure: 12 mLength of structure (1): 15 mMaximum permeability of building: 3 percent of wall area
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sP64(s&T):2001
Wind Data
Wind zone : 2 (Basic wind speed= 39 m/s)Terrain category :2Class of structure :AProbable design life of structure : 50 yearsTopography : Almost flat
Design Wind Speed
VZ = V~.kl.kz.k~Vb = 39 dskl = 1.00kz = 1.00k~ = 1.00Vz =39xlxlxl=39m/s
Design Wind Pressure
p,= 0.6(VZ)2 = 0.6(39)2 = 912.6 N/m*
Internal Pressure Coeflcients: C~i = * 0.2
External Pressure Coeflcient on Roof [6.2.2.10 and Table 20]
bl=5m, bz=2mthatis, bl>bz
For e = 0°CP, for portions a and b is –0.6 and +0.7 respectively,
For 6 = 180°The values of CP, as obtained for Et= 0° get interchanged.
External Pressure Coeflcient on Walls (CPJ [6.2.2.1 and Table 4]
Mws 0.5 and l/w= 1.25
.
Region A B c D Local
When9=0° +0.7 -0.2 -0.5 -0.5 I -0,8e =900 -0.5 -0.5 +0.7 -0.2
CPLocal Values for Rooffrom [Table 5], for (3= 0°
Region a d e fgh
Cp~ when Cpi= +0.2 -0.6 -0.5 -0.8 -2.0 -2.0 –2.0
Cp(net) when Cpj= –0.2
CD(net)
-0.8 -0.7 -1.0 -2.2 -2.2 –2.2
-0.4 -0.3 -0.6 -1.8 -1.8 –1.8
CP(net) for roof are shown in Fig. 22 (a) - (d),
CP (net) for Wall
CP(net) Local= – 0.8 – 0.2= –1.0
CP(net) for main portions of the wall
Region A B CD Local
o = 0° and C’pi= + 0.2 +0.5 -0.4 -0,7 -0.7 -1.00 = 0° and cpj = -0.2 +().9 +-0.0 -0.3 -0.3
..
I
,,
/
67
I
SP 64 (S & T) :2001 L ______ ~~,.
1Region A B CD Local
0 = 90° and Cpi= + 0.2 -0.7 -0.7 +0.5 +0.4 ]
J -1.0(3= 90° and C~i= – 0.2 -0.3 -0.3 +0.9 +0.0Design
1
+0.9 -0.7 -0.7 –0.7Pressure -0.7 0,0 -0.9 0.0 1
–1.0
Coefficient
Figure 22 (e-h) give the values of coefficients for the main walls.
Since ~ and Ub are less than 4, there will be no frictional drag on building.
NOTE—Thatalthough‘Local’valuesatthecomerofroofsandwallsarenotspecificallyindicatedin[Table20,Tables4and5]shallbe used in the case of all buildings for localvalueson the roofs and vertical walls.
Design pressure coefficients for the whole building are given in Fig. 22(i).
e
~,
bCd -Ja 0$ M
76.Om
J
=5m 2 b! ● 5m
ELEVAf ION
1 12.0 m 1.r
9 1
ROOFPLAN (LOCAL)
(a)
b = 0.7 48
c1
a = -0.6 -0.6 ~c5
(b)
-0.8 ba +.0.7
0
a= -tM
0=180”
(c)
Cp FOR ROOF
-1.0 +09
o
+0 -08
(d)
Cp NET FOR ROOF
I
.’
FIG.22A RECTANGULARCLADBUILDINGWITHSKYLIGHTROOF(Continued)
68
II ~
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SP64(S&T) :2001
c -0.7
1’
c -0.3
A s A B
~ lS.0m4
+0.5 -0.11
,1 ,
+0.9 0.0
II -0.7 0 -0.3
%t2.Om 12.Om \
wHEN 0=0*, CPI = +0.2 WHEN 8 = 09, Cpi = ‘0.2
(e)
c +0.t
A ‘5’181
-0.7 +7
—D -0.6
*12.Om Y
wHEN S=9Cf, Cpi = +0.2
(9)
Am
-0.3
WHEN 9 = 9U, CPI=-0.2
(h)
3
.0.3
22 (e-h) ~(net) for Main Walls
FIG. 22 A RE~ANGULARCLADBUILDINGWITHSKYLIGHTROOF (Continued)
/
!,’
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SP 64 (S & T) :2001
-1.0 ( LOCALON WALLS)
+0’91-04
VERTICALWALL ‘b’
Zzz+09, -0.7
Zz
WALL
-04
[
-0-s-1.0
i
ROOF
Zzzz+0-9 ,-07
E%%
WALL
LS)
A
22(i) ~(net) for the Building
FIG.22 A RECTANGULARCLAD BUILDINGWITHSKYLIGHTRGGF
Example 8. Consider the rectangular clad building of Example 7 with the same parameters except thatbl = b~ (Fig. 23) [Table 20].
In this case bl S b2, therefore, values of CP for portion a and b are to be determined from [Table 20].
Here bl = bz and hl/h2 = 6.8/6 = 1.13
CP~(for portion a) = 2 x 1.13-2.9 = 4.64and Cpe (for portion b) = -0.5 (for direction 1)
Cpe(for portion b) = -0.4 (for direction 2)
CP Local values for roof from [Table 5], fore= 0°
:,,-
i’
Region a b e c d
Cpe when Cpi= +0.2 -0.64 -0.5 -0.8 -0.6 -0.5
Cp(net) when Cpi = –0.2 -0.84 -0.7 –1.0 -0.8 -0.7
Cp(net) -0.44 -0.3 -0.6 -0.4 -0.3
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SP64(S&T):2001
CPLocal values for roof from [Table 5], for9=180°
Region a b e c d
Cpewhen Cpi = +0.2 -0.5 -0.6 -0.8 -0.5 -0.64
Cp(net) when Cpi = –0.2 -0.7 -0.8 -1.0 -0.7 -0.84
Cp(net) -0.3 -0.4 -0.6 -0.3 -0.44
Cpdesign values -0.84 -0.8 –1.0 -0.8 -0.84
However, on leeward side as per [Table 20, Item ( b)], the Cw for pitched portion ‘a’ and vertical portion ‘b’are -0.5 and -0.6, respectively Fig. 23 (b) and (c).
CP(net)values considering variability of wind direction are given in Fig. 23 (d).
Other values of CP(net) as well CPlocal remain unchanged.
CP(net) for the roof and the walls are indicated in Fig. 23(e).
ELEVATION
(a)PLAN
e.(y
(b)
Cw FOR ROOF
r
-* -w -0aw
:\,
.-
0=180’(c)
L
(d)
Cp NET FOR ROOF
FIG.23 A RECTANGULARCLADBUILDINGWITHSKYLIGHTRGGFSPECIALCASE(Continued)
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SP 64 (S & T) :2001
-1.5 (LOCALON WALL)
r -1.5(~A ON WALL)
0.8
VERTICALWALL’G
484
R0“0F
-oa4
A--
L3.Om 3.Om12.0m
7 1
23 (e) ~(net) for the Building
FIG.23 A RECTANGULARCLADBUILDINGWITHSKYLIGHTRGGFSPECIALCASE
r
r
/
72
I
I I11
py. ‘.. .
-,, ● LA*——
..
SP 64 (S & T) :2001
Example 9. Consider the rectangular clad building of Example 9.1.2.7 with the same parameters except that
the skylight is of triangular shape (Fig. 24) [Table 20].
The Cpeon roof are given for all roof portions as for pitched roof and skylight portion as –0.6 and +0.4 forwindward side and –0.6 and –0.5 for leeward side.
Thus CP(net) considering the variability of direction of wind is -0.8 and +0.6 for all portions of ,tie roof andCP(net) as well as CPlocal remain unchanged for other portion of building.
I-f
--&
n.a bcd
760m
J-
1-bt :5m ! 1< b,. 5m
“’bf21+ -1
ELEvATION
(a)
b =+011 c*-O.6
m
a=-0.6 d:-o. ~
0=0”
(b)
Cpe
+ 120m —~ROOF PLAN (LOCAL)
CZ-O.6 b.+04
n
d=-O.5 •--o.~
e = 160”
(c)
FOR ROOF
I(d)
Cp NET FOR ROOF
FIG.24 RETANGULARCLADBUILDINGWITHSKYLIGHTRGGF
Example 10. A godown with semi-circular arch springing from ground (Fig. 25) [Table 15].
Given:
Physical Parameters:
Height (H)Width (1)Length (L)RoofPermeabilityExternal surfaceTopography
Wind Data
Wind zone :Terrain category :Class of structure :
5.0 m10.0 m30.0 mSemi-circular curved roof, springing from ground levelOpenings on sides not more than 5 percent of wall areaCorrugatedAlmost flat
4 (Basic wind speed= 47 m/s)2,B
. . ..—.— $
.;
I
iI
,,,.
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SP 64 (S & T) :2001
Design Wind Speed
VZ = Vb.kl .kz.k~v~ = 47 In/skl = 0.90 (for mean probable design life of structure as 25 years) [Table 1]kz = 0.98 [Table 2]k~ = 1.00 (for flat ground) [5.3.3.1]VZ = 47X 1.00X 0.98X 0.90= 41.45 IU/S
Design Wind Pressure
p,= 0.6 x (VZ)2 = 0.6 (41.45)2 = 1031.0 N/m2
Internal Pressure Coetllcient [6.2.3 and 6.2.3.1]
Since openings are not more than 5 percent of wall area C~i= * 0,2
External Pressure Coeftlcient for Roof [6.2.2.5, Table 15 andTable 5]
For H/l = 5/10 = 0.5 ; C= –1.2 and Cl = + 0.7
The curved roof is converted to pitched roof at 3.5 m height upwards and coefficients for this portion arecalculated as per [Table 5] as shown in Fig. 25(a).
For 6 = 0° and a = 22.6°
EF GH Local 1 Local 2
Cpe -0.3 -0.4 –1.0 –1.2Cp(net)if C~i= +0.2 -0.5 -0.6 -1.2 -1.4if Cpi= -0.2 -0.1 -0.2 -0.8 -1.0
Design Cp -0.5 -0.6 –1.2 -1.4
For Et= 90° and a= 22.6°
EG FH
Cpe -0.7 -0.6CP(net)if Cpj= +0.2 -0.9 -0.8if Cpi= -0.2 -0.5 -0.4
Design Cp -0.9 -0.8
CP. (local) are indicated in Figs. 25 (b) and (c)
External Pressure Coeftlcienta for End Walls
h/w= H/1=0.5 andl.5<Uw=l/L<4.O
Wind angle c D Local
Cpefor 0 = O“ -0.6 -0.6 –1.0
CP when Cpi = +0.2 -0.8 -0.8 -1.2
Cp when Cpi = –0.2 -0.4 -0.4 ~.8
Cp. for 0 = 90° +0.7 -0.1 –1.0
Cp when Cpi = +0.2 -0.5 -0.3 -1<2
CP when Cpi = –0.2 +0.9 +0.1 -0.8
,,>.
Width of local edges= 0.25 w or 1= 2.5 m
CP(net) values for roof are shown in Fig. 25 (d).
74
SP 64 (S & T) :2001
Frictional Drag [6.3.1]
h= H=5m, b=l=lOm, d= L=30m,
pd = 1031.0 N/m*
Since W >4 and W/I< 4, the frictional drag will be experienced by structure for (3= 90° and 270°. Also since,the roof is curved there are no walls. The frictional drag is given by:
F’= C{ (d – 4h) b. Pd. and C; = 0.02 for corrugated surfaces.F’=0.02(30 -4 X5) X1 OX1O31.O=2M2.ON
Half of the frictional drag is to be resisted by windward l/3rd (10 m) length and rest half by 2/3rd (20 m) lengthon leeward side, in addition to other loads.
Forces
F= CP.A.p~ and pd = 1031.0 N/m*
Cp Force,N/mz
4.9 -927.9
+.8 -824.8
-0.6 -618.6
4.3 -309.3
-1.2 -1237,2
-1.4 -1443.4
,(CQ-t--t5.
-d35m
10m
CPQ FORROW(0-Cf’l(b)
a.“la’ “\--t
YJm
-1
(a)
[ ;:;’., -0.5
k !
“i -i
------ “______
5m ‘ -0.7
Cp, FOR RUM (tih9d’1
(c)
/
FIG.25 SEMI-CIRCULARSHELLROOF(Continued)
,
75
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SP 64 (S & T) :2001
(Q!%-t36m1.2 ‘
LOCALl’2
-115m lm 15
1 c +0.9,
-w
~~~1
-1.2
-1.2-0.6
30mA 4
L 10m 1
Cp (NET)WHEN 9=0”
T--i--T
CP(NET ) 9=9$
FIG. 25 SEMI-CIRCULARSHELLRGGF
Example 11. Rectangular clad structure having circular arch shell roof with overhang (Fig. 26) [Table 15].
-—.
7H.3m
o(= 55°
OVERHAIW
TIE~1 TIE
J
m
Given:
L 2.5 m 1 i-11 m 1 2.5 m L’-1 ~ 1 1
Physical Parameters:
FIG.26 BUILDINGm CIRCULARARCHSHELLRGGFWITHOVERHANG(Continued)
1’
,1
Height of arch roof (H) : 3mHeight of structure : 6mWidth (1) : llmLength (L) : 30mRoof slope : 65°
76
/
RoofOpenings on sidesExternal surface of cladding :
Topography
Radius of shell
Wind Data
Wind zoneTerrain categoryClass of structureMean probable life of structure :
Design Wind Speed [5.3]
VZ = V~,kl.k2.kJkl = 0.90kz = 0.98k~ = 1.00
SP 64 (S & T) :2001
Curved circular archOpenings 5-20 percent of wall areaRoof made of ferro-cement flat units overlapping each other withdimensions0.8 m x 2 m; masonry (unplastered) for walls.Almost flat with less obstructions in surrounding but industrial areafar away6.54 m
4 (Basic wind speed= 47 m/s)2B25 yearn
v= = 47 X0.90X 0.98X 1.00= 41.45 m/S
Design Wind Pressure
pZ = 0.6 (VZ)2 = 0.6 (41.45)2= 1031.0 N/m2
Internal Pressure Coefficient [6.2.3 and 6.2.3.2]
Since openings are about 5-20 percent of wall area, the value of CPi=* 0.5. The sign will depend on the directionof flow of air relative to the side of openings.
Pressure Coefficient (on roof) [6.2.2.5, Table 15 andTable 5]
Here HA= 3/1 1 = 0.273 and 0.7H = 2.1 m
}~ ‘=-’09~~8 (by interpolation)
2’
For the purpose of local C and CPi,the arch roof is considered of equivalent pitched roof havinghAv=6/11 =0,545 and roof angle= 65ge
Since the ridge is curved there will be no local coefficients. The Cm values for the roof are shown inFig. 26(a) to (c)
Pressure Coefficient for Overhangs [6.2.2.7]
CPeat top of overhang portion is the same as that of the nearest top portion of the roof and in this case Cw for0 = 270° and 90° is -0.7 for 5.5 m width at edges and –0.5 for middle strip of 2.45 m.
For windward side of overhang the equivalent CPi(on under side of roof surface) is 1.25 while on leeward sideof overhang the CPiis equal to the CPeof adjoining wall.
CP(net)values forthe roof are indicated in Fig. 26(d).
External Pressure CoetTicient for End Walls [6.2.2.1 and Table 4]
It/w = 6/11 = 0.54 and VW= 30/11 = 2.73
Wall A B c D Local
for Q=0° +0.7 -0,3 -0.7 -0.7
}–1,1
for 8 = 90° -0.5 -0.5 +0.7 -0.1
.,
Comer width is 0.25w = 2.75 mCP(net) values for the roof are indicated in Fig. 26(e).
77
SP 64 (S & T) :2001
Frictional Drag [6.3.1]
h= If=3m, B=l=llm, p~=1031.0N/m2d= L=30m
Since fi = 10>4 and Wh = 3.67<4, frictional drag will be experienced by structure for (3= 90° and 270°.
~ = C; (d - 4h)b.p~ + Cr (d – 4h)2h.p~
Ct’ = (average of smooth and corrugations surfaces)= 0.5(0.01+ 0.02)= 0.015
F’= 0.015(30–4 X3) X11 X1 031+0.015(30– 4X3) X2 X3 X1O31= 3062.1 N+ 1 670.2N
(roof) (wall)
Half of frictional drag is to be borne by windward l/3rd length that is, 10 m length and rest half of frictionaldrag by the 2/3rd length on leeward that is, 20 m, in addition to other loads.
Forces
p~ = 1031.0 N/m*
Cp Force, N/m*
–1 .75 -1804.25
-0.8 -924.8
I 0.9 I 927.9 II -1.2 i -1237.2 I
-1.0 -1031.0
-1.473 -1518.66
-9,08 -936.15
-1.6 -1649.6 Ic= -0..973
-0.
WHEN O=0°
(b)
-Q,973
-o-b
WHEN 9=270°
(c)
26 (a) to (c) ~ on RoofFIG. 26 BUILDINGWITHCIRCULARARCHSHELLRGGFWITHOVERHANG(Conthwe@
78
I
SP64(S&T):2001
II
I
-0.8
26(d) ~(net) for Roof
A+1.2-1”0 I
I
26(e) ~(net) on wall
0=270°
B+1.2-1”0
FIG.26 BUILDINGWITHCIRCULARARCHSHELLRGGFWITHOVERHANG
,,
,.,
79
SP 64 (S & T) :2001
Example 12. Estimation of design wind pressure on glass glazings.
The determination of equivalent static wind pressure on large glazing has been the subject of considerable studyever since the failure of such glazing on the Kennedy Memorial Building in Boston, USA. It is recognised thatsuch large glazing experience three types of wind loads, all additive, namely:
(i) a static wind load due to a wind speed averaged over a specified time interval,
(ii) a random dynamic wind load due to turbulence in the wind, and
(iii) a resonant wind load due to resonant excitation of the vibration of the panel by the energy of theturbulent fluctuations at the natural frequency of the panel.
Generally speaking, the resonant component can be neglected if the natural frequency of the panels is more thanabout 250 Hz. For design purposes, the information from experimental and theoretical studies have been codifiedinto equivalent steady design wind pressure, in which all of the three wind load components stated above arealready added appropriately.
Although the design wind pressure can be worked out for each level of a building, it is recommended that thepressure corresponding to the top level as well as those at the comers be adopted for all lower levels. There aretwo main reasons for this recommendation:
First — The assumed need to standardize the thickness of the glass at all places.
Second—The effect of failure of a panel at any height. Such failure may increase the internal pressure coefficientto positive like 0.8 due to free passage of air into the building as against the maximum value of 0.2 for buildingwith no windows. Such increase of internal pressure may bring about the failure of other panels to which thehigh internal pressure may be communicated by the passages inside the building.
To illustrate, take an example of a square cross-section sealed building just outside New Delhi of height 50.0 mand sides of 15.0 m each, in terrain category 2 and with no windowslopenings.
Design Wind Speed [5.3]
VZ = V~.kl.k2.k~k, = 1.07, since failure of glass panel is hazardous [Note 2 of 6.2]kz = 1.17, though the size of building is 50 m but the panel size will be less than 20 m, hence the panel is
considered as Class A and corresponding value has been picked up from [Table 2]kx = 1.00VZ = 47x 1.07x 1.17x 1.00 =58.84 m/s
Design Wind Pressure
pZ = 0.6 (VZ)2 = 0.6 (58,84)2=2 077.24 N/m2
Since the building is without windows, as per [6.2.3.1], the internal pressure coefficient=* 0.2.
Also, (3/2) c (h/w)< 6 and lC (Vw) c (3/2)
Considering that glazing is provided at a distance of less than 0.25 times the width of wall from the comer, asper [Table 4] the external local pressure coefficient of –1 .20 will govern the design even with minimum internalpressure of+ 0.2; resulting in a net pressure coefficient of –1 .4. This gives a design pressure for glazing
p~=2077.24 x 1.4=2 908.13 N/m2
9.1.3 Preliminary Explanation with Regard to Structures with Free Roofs [Covered by Tables 8 to 14]
[Tables 8 to 14] refer to roofs with open bays on all sides, [Table 8] applies to roofs with (Ww) lying between0.25 and 1.0 and (L/W) lying between 1 and 3. Modifications to the roof pressures due to obstructions are to beapplied. When the length of the canopy is large, then [Tables 9 to 14] have to be used, if (b/d) = 5.0 or nearabout. In such cases [Tables 9 and 10] imply that the average pressures will have to be worked out for portionsof the roof for wind angles of 0°,45°, 90° and 0°,45°,90°, 135° and 180° when there is obstruction on one side.For example [Tables 9 and 10] give separate pressure on one quarter leeward area of roof, designated as J, at6 = 45°. [Tables 11 and 12] are to be used if the canopy is inclined at 30°. In case (b/d) is substantially morethan 5.0 and roof angle is other than those given in the Code, it is better to commission wind tunnel model studies.
,,. .
i’
Frictional force shall be considered when the roof area is larger in [Tables 9 to 14], compared to the area in[Table 8]. The friction force is to be distributed such that half of the friction force acts on the windward l/3rd
80
1SP 64 (S & T) :2001
1 of the surface and the other half on the leeward 2/3rd surface. The forces may be considered to act at the midplane of these portions.
It The definition of the height ‘h’ given in [Table 8] for positive and negative roof angles must be carefully noted.It is not the height to the lowest point of the canopy for both positive and negative roof angles.
I
The definition of the width is that of the main canopy only and will not include the facia if they are inclined andIextend beyond the edge of the main canopy.
iCase A — No obstructionw’stored material below the canopy
I[Table 8, read with 6.2.2.4], gives the overall roof pressures of the canopy. The columns, trusses and foundationsIare to be checked for these loads. It is to be noted that the Code gives only the force coefficients in the direction
I of wind as 1.3 for the facia implying that only a facia of angle 90° (that is, a vertical facia) is covered in theCode. If the facia has an angle other than 90°, the loads can be significantly higher and values based on wind
rtunnel model studies or conservative values based on the loads of roof overhangs of buildings [6.2.2.2 and Table5] maybe used. If a number of such canopies are being built, it is more economical to get model studies madein wind tunnel to know accurate loadings.
Case B — Obstructions/stored material below the canopy
Here, the solidity ratio as defined in [6.2.2.4] has to be found. This is obviously the ratio of the height ofthe obstruction to the total height of the canopy. As indicated in [6.2.2.4 and Table 8], the positivepressure is not affected by the obstruction but the negative pressure upstream of the obstruction getsmodified. The horizontal force coefficient of 1,3 on the vertical canopy is introduced only on thewindward canopy.
The pressures at the corners A, B, C, D are shown at the higher value of –3,0 in line with some of the new findingsand in the absence of a specific clarification on this point in the Code. It may be noted that as per the Code, thedesign cannot be checked for opposite pressures on the two arms on the canopies also.
i9.1.4 Solved Examples of Free Roofs
I Example 13. A rectangular clad building with free pitched roof [6.2.2.4 and Table 9].
t Given:
Physical Parameters:p
Height (h) : 5mI Length (b) : 50m~ Span (d) ; 10m
Roof angle (et) : 30°
tWind Data
1 Wind zone 4 (Basic wind speed= 47 rrds)Terrain category 1
I Class of structure BI Design life of structure 25 years.I
1 Design Wind Speed
V,= Vb.kl.k2.k3 = 47 X 0.90 X 1.03X 1.0 = 43.57 tllk+
Design Wind PressureI
pZ = 0.6 (VZ)2 = 0.6 x (43.57)2= 1139.00 N/m2
CASE I: When there is no stored material underside [see Fig. 27(a)l
‘........-.....~...z,:,::
,,~>.
Wind Load Calculation
Pressure Coeftlcient on Roof and End Surfaces [Table9]
81
a. -.,
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SP 64 (S & T) :2001
For Roof
CP(net) = CP(top) – CP(bottom)
For (1= 0°CP(for region D) = 0.6 – (–1.0) = 1.6CP(for region@= -0.5 – (–0.9) = 0.4
For 0 = 45°CP(for region D) = 0.1 – (-0.3) = 0.4CP(for region E) = -0.6- (-0.3) = -0.3CP(for region J) = -1.0- (-0.2)= -0.8
For Q= 90°CP(for region D) = –0.3 – (–0.4) = 0.1CP(for region@= -0.3 – (–0.4) = 0.1
For End Surfaces
CP(net)= CP(front) – CP(rear)CP(windward, end surface)= 0.8 – (0.4) = 0.4CP(leeward, end surface)= -0.3 – (0.3) = -0.6
For f3= 90°, there will be force due to frictional drag (Tangentially)= 0.05p#d= 0.05 X 1 139.00X50X 10=28 475 N
Force on portion D = 1.6x 1 139.00x 1 = 1822.40 N/m2
Force on portion J = 0.8 x 1 139.00=91 1.20 N/m2
Force on portion E = -0.3 x 1 x 1 139.00= –341.70 N/m*
The design pressure coefficients are indicated in Fig. 27(b)
CASE II: If effects of train or stored materials are to be considered [see Fig. 27(c)]
Wind Load Calculations
Pressure coefficients on roof and end surfaces are to be taken from [Table 10]
For 0 = 0°CP(for region D) = 0.1 – 0.8 = -0.7 for full lengthCP(for region E,)= -0.7 – 0.9 = –1.6 for full length
For 0 = 45°CP(for region D) = –0.1 – 0.5= -0.6CP(for region q = -0,8 -0.5 = –1.3CP(for region J) = -1.0 -0.5 = –1.5
For 0 = 90°CP(for region D) = –0.4 + 0.5 = 0,1CP(for region E) = -0.4 + 0.5 = 0.1CP(windward, end surface)= 0.8+ 0.4= 1.2CP(leeward, end surface)= -0.3- 0.3= -0,6
For e = 180°CP(for region D) = –0.3 + 0.6 = 0.3CP(for region E) = 0.4 + 0.6 = 1.0
Design pressure coefficients are indicated in Fig. 27 (d).
For (3= 90°, Frictional tangential force = 0.05 pdb.d
Forces on Different PortionsF1 = 1.2x 1 139.00= 1366.80 N/m2Fz = –2.0 x 1 139.00= –2 278.00 N/m*F~ = –1.6 x 1 139.00= –1 822.40 N/m2FA= –0.7 x 1 139.00= – 797.30 N/m*
82
SP 64 (S & T) :2001
t’ II I
“’l-y--l
Ll_IE
. ..--.. —-.Lddom
27(a) Plan
-0.6$ +0.41
r -
1
+
--f~+ 1.6, .9 d
1
b/3
\- 1
---- -----
b +1.6, - .3
[ : : -
b/3
---- -----t
,4.J.S TJ
b/3a
A-0.6, +0.4
DES!GN PRESStJiE C&F ICIENT
27(b)
ROOF SLOPE X = 30-
EFFECTS OF TRAINS OR STOREDMATERIALS0= O“ -45”, OR 135”-180”
D, D, E, 1? FULL LENGTH
e= 9w, D, D. E. F PART LENGTH b,
THEREAFTER OP = O
F/2
F/4
F/2
FRICTIONAL DRAGDISTRIBUTION
FIG. 27 EXAMPLEOFPITCHEDFREEROOFS,ct = 300m ORWmHOUTErwmrs OFTRAINORSTOREDMATERIALS(Continued)
I
1
,,,,
,,’
83
/
..
SP 64 (S & T) :2001
+--l
K-=--l
27 (C)
t
r‘“ -f~-2.0 d
-L
W3
‘!” ,+
-.. ---
b+~.o.7
L ‘T: ‘
b/3
---- -----
7
tW2
~ J
d
-i
ROOFSLOPEX=‘WEFFECTS OF TRAINS OR STOREDMATERIALS9= W-4Y,0R135” .180”
D, D, E, E’ FULLLENGTH0 = 90”,0, D, E, E’ PARTLENGTH b’
THEREAFTERCP= O
FI 2
F/4 FRICTIONAL ORAG
DISTRIBUTION
$12
27 (d) Design Pressure Coefficient
FIG.27 EXAMPLEOFPITCHEDFREEROOFS,et = 30° wrim ORwrrHouT Emo3crs OFTRAINORSTOREDMATERIALS
Example 14. A free double sloped tempora?y roof (Fig. 28) with and without an obstruction underside, andwith and without a facia of 0.75 m [6.2.2.4 and Table 8].
Given:Physical Parameters:Height of roof(h) : 3.41 m
Width of roof (w) : 5.0 m
Length of structure (1) : 10.0 m
Roof angle (CL) : 20°
The structure is supported on columns at a distance of 1.0 m from the centre. There are 1.0 m high cotton balesalong its length, with width of 1.0 m.
,.,.,
84
SP 64 (S & T) :2001
scorn L“ -
h.sam
1
I
r-2.5m
-1- mn
ELEVATION
$5m HIGH WAL
/f//l/fl//~~~~f~f
t333 m L 3.33 m L 3933m
1 1 “F”’”; J
t—-+- --t-”- 5.0 m
-~lo.om~1
PLAN
FIG.28 A FREEDOUBLE-SLOPEDROOF(Continued)
Wind Data
Wind zone
Terraincategory
Class of structure
Life of structure
Topography
Design Wind Speed
Vz = V@l .k2.k3
v~ = 39 n-h
kl = 0.76
kz = 1.0
k3 = 1.0
Vz = 39x0.76 xlxl=29.64ds
..
..- .. .. . ..
t
f$
>
$,j!
4
:;\..,’
,
%
..2~.-.
2 (Basic wind speed= 39 m/s)
2
A
5 years [Table 1]
Flat
/
85
1
I .’SP 64 (S & T) :2001
Design Wind Pressure
p== 0.6 (VZ)2 = 0.6 (29.64)2= 878.53 N/m2
Pressure Coefficients on Roof
CP for O= 0° [6.2.2.4 and Table 8]
Since the 1.0 m high cotton bales are normal to wind direction and is on upwind, the solidity ratio(+)= 1.0/2.5= 0.4
Maximum +ve (thmst)
Cp(overall)=+ 0.7Cp(local)= +0.8, +1.6, +0.6, +1.7 [As shown in Fig. 28 (a)]
Maximum –ve (suction)CP(overall)= -0.78CP(local)= –1.08, -1.54, –1.78, -0.96 [As shown in Fig. 28 (a)]
CP for 0 = 90°
Since there is no obstruction for air flow $ = O, All the thrust values of CP are same as in the case for 6 = 0°
Suction values are
CP(overall)= –0.7CP(local)= –0.9, –1.3, -1.6, –0.6
CPfor 0 = 180°
Since the obstmction is on downwind side not on upwind side 1$= O.The CPvalues are same as in the case ofe = 90°.
The design pressure coefficients CPare shown in Fig. 28 (a).
Force Calculations
The force on ridgeltrough induced by individual slopes is as follows:
F = Cp(overall) x A x p~ (where A = Area of one side roof)
= –0.82 X 2.5X10X 878.53= –18 009.86 N
F= +0.7 (2.5x 10)x 878.53= 15374.28 N acting on centre of roof,that is (2.5/2 = 1.25 m) away from the trough, acting normal to the surface
Total upward force on support structure= 2 x (-18 009.86)=–36019.72 N = -36.02 kN (upward)
Total downward force
F= 2 X (15 374.28)=30 748.56 N = 30.75 N
Maximum moment induced on support structure is when one side roof experiences thrust and other side roofexperiences suction.
Hence the moment at top of support due to vertical component of wind= (-18 009.86X 1.25-15 374.28X 1.25) COS20°= -41729.1 cos 20°=39212.53N-m
4-....—– “
L....
+..----
.
~
Moment at the base of the support due to horizontal component of wind(18 009.86 + 15 374.28)x 2.5 sin 20°=28 545.12N-m
86
t$ mL
RI
—0”8 , -1c08~1”7, -o*%
%
1.6,-1054
0.7,-0.82
H
v
096, -1*78
0.8 & 1.78 (
Cp LOCAL
28(a) Forces on Support Structure
FIG.28 A FREEDOUBLE-SLOPEDROOF
SP 64 (S & T) :2001
HIGHER OF TWOFOROVERLAP)
Example 15. A rectangular unclad structure with V- troughed free roof (a = 10’) aria’underside obstruction(Fig, 29) [6.2.2.4, Tables 13 and 14].
Given:Physical Parameters:Height (h) : 4mWidth (d) : 8mLength (b) : 40mRoof angle (a) : 10°
Since the inclination of the canopy is 1(Y [Tables 13 and 14] have to be used.
Wind Data
Wind zone 2 (Basic wind speed 39 rrds)Terrain category 2 [5.3.2.1]Class of structure B [5.3.2.2]Life of structure 50 years.Topography Flat
Design Wind Speed [5.3]
VZ = Vb.kl.kz.k~v~ = 39 mlskl = 1.0kz = 0.98k~ = 1.0 (for flat ground)v, = 39X 1 X0.98X 1 = 38.22 m/s
Design Wind Pressure
...-—---
1
-- —-’.-. ..
f .,:
~’
P.= 0.6 (V,)2= 0.6 (38.22)2 = 876.46 N/m2
87
/
SP64(S&T):2001
Without underside obstruction. [Fig. 29 (a)]
Pressure Coefficients [Table 13]
On roof CP(net)= CP(top)– CP(bottom)
Positive forces on a surface are always to be taken as acting towards the surface.
For (3= 0°
CP(net) (for region D)= CP(D)- CP(D’)= +0.3 – (-0.7) = +1.0 on full length of roof acting downwards
CP(net) (for region E)= 0.2 – (–0.9) = +1.1 on full length of roof acting downwards
For 6 = 45° (on full length of roof)
CP(net) (for region D) = 0.0- (-0.2) = +0.2
CP(net) (for region E) = 0.1- (-0.3)= +0.4
For O= 90°
CP(net) (for region D) = -0.1- (0.1)= -0.2
CP(net) (for region E) = -0.1- (0,1)= -0.2
Acting on windward width (b’= d ) and is zero on rest of portion of roof. [6.2.2.4 and Table 13]
CP(net) (for region F ) = 0.4- (-1.5)= +1.9 [Fig. 29 (a)]
Effect of obstructions below the roof [Table 14]
Pressure Coefllcient on Roof
Cp(net) = cp(top) - CP(bottom)
For (3= 0° on full length
CP (for region D) = -0.7- (0.8)= -1.5
CP (for region ,/7)= -0.6 – (0.6)= –1.2
CP (for region F) = -1.1- (0.9) = -2.0
For 0 = 45° on full length
CP(for region D) = -0.4- (0.3)= -0.7
Cp(for region E) = -0.2 – (0.2)= -0,4
For (3= 90° for windward length of b’= 8,0 m
CP(for region D) = -0.1- (0.1)= -0.2
CP(for region E) = -0.1- (0.1)= -0.2
For 6 = 180° on full length
CP(for region D) = -0.4- (-0.2) = -0.2
CP(for region E) = -0.6- (-0.3) = -0.3
Friction force at 0 = 0° and 180°
= 0.1 p~.b.d= 0.1 X 876.46X40X8= 28.04 kN
CP(design) for cladding
Since obstruction may be on either side, hence, maximum CPis –1.5 over the roof and CPlocal on edges is -2.0.Recent measurements suggest that the value should be enhanced to –3.0 at the corners.
88
1
I& ‘T
UTd+—d --------
t- —-- ---
Cp=l”.
D E
~; J
~=d
e C,=f.1
u c,=l.o~
4RgO”
b=5d
I
LfiJl=0.2d
---- ---
d=8
CASE e =0’
RIDGE
IIJJ.Cp-o-o
CASE e =90°
SP 64 (S & T) :2001
~cP =0”2
p=o.1+
~
CASE (3=45s
29 (a) ~(net) Design for Cladding with Local Coefficients on Ridge
t=d
i
~ ml
1=d
Cp--0.7
Cp= -04
I CASE e=tf CASE e =45’
q [
-2.0Cp(u)
Cp=-o.ltb
[u
-1.5
. . .——.-,
_ ..—
~
I,+
$
I
i
$
CASE 9 =96’ CASE 0 = 45”
29 (b) ~(net) Design for Cladding with Local Coefficients on Ridge with Underside Obstruction
FIG.29 CP(net) DESIGNFORCLADDINGWITHORWITHOUTUNDERSIDEOBSTRUCTION
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9.2 Force Coeftlcient Method [6.3]
9.2.1 In the Force Coefficient method the total wind load F on a particular building or structure as whole isgiven byF= Cp A.. p~whereCf = force coefficient,A,= effective frontal area of building or structure, and
p~ = design wind pressure.
The Code specifies, different values of force coefficients for different faces of a building. Since the designpressure varies with height, the surface area of building/structure be sub-divided into strips and appropriatepressures are taken over the strips and calculated individually on the area Ae.
9.2.2 Force Coefficients for Clad Buildings [6.3.2]
[Figure 4] gives the values of the Force Coefficients for rectangular clad buildings of uniform section with flatroofs depending upon the Mb and a/b values while [Table 23] gives the value for other clad buildings of uniformsection.
9.2.2.1 Force coefficients for circular sections [6.3.2.2]
Though the Code specifies that Force Coefficients for circular cross-section buildings be taken from [Table23], yet for a precise estimate for infinite length [Fig. 5] may used which accounts for the surface roughness ‘E’and in case the length of the structure is finite the Force Coefficient is modified by a multiplication factor of K.
9.2.2.2 Force coeficientsfor walls and hoardings
Earlier it was thought that the maximum wind pressure occurs when the wind direction is at right angles to thesurface. The wind tunnel tests abroad, however, indicated that the maximum net pressure occurs when the windis blowing at approximately 45° to the wall/hoarding. Further, as the width to height ratio of wall increases, thenet normal pressure across the wall, for wind direction normal to the wall, decreases due to the attenuation ofthe pressure fluctuations on the leeward side. Near a free end or corner, there is a large increase in load foroblique wind directions blowing on the free end specially at higher width/height ratios. In case, an adjacent wallis running at right angle to the free end the net load near the corner reduces considerably.
In case of hoardings the pressure tends to be a little higher due to the wind flow under the hoarding. When thegap underneath the hoarding. The Code, therefore, recommends in [Table 24] the value of Cffor hoardings/wallsofs 15 m with Vwious bfi ratios, if resting on the ground or above tie groundataheight20.25 h (b and h are
the width and the height of the wall/hoarding respectively).
9.2.3 Force Coefficients for Unclad Buildings
The Code [6.3.3] specifies the values of Force Coefficients for permanently unclad buildings and for frameworks of buildings which are temporarily unclad like roof trusses and frame works of multistoreyed build-ings/factories during erection.
9.2.4 Force Coefficients for Individual Members
For individual members of infinite length the Force Coefficients are to be taken from [Table 23] and for finitelength members depending upon the Z/dor Ubvalues to be modified by a multiplication factor K as before. Thesevalues are further modified in case the member abuts onto a plate or wall in such a way that the free flow of airaround that end of the member is prevented. In such cases the Ubvalue is doubled for determining the value ofK. In case the free flow of air is prevented around both ends of the member the value of Ubis = m for calculatingthe value of K.
9.2.4.1 Flat sided members
[Table 26] specifies the values of the Force Coefficients for two mutually perpendicular directions in case offlat sided members when the wind is normal to the longitudinal axis of the member. These are designated as Cf.and Cftrespectively, for normal and transverse directions and are given in [Table 26]. The normal and transverseforces are thus given by the following two relations:
,,. .
Normal Force, F.= C~n.P~.K. 1. b
Transverse Force, Ft = Cfr pd. lC.1. b
90
i’!
SP 64 (S & T) :2001
9.2.4.2 Circular sections
The values of Force Coefficients for such members are to be calculated as explained in 9.2.2.1.
9.2.4.3 Wires and Cables
In the case of wires and cables the values of Force Coefficients are dependent on diameter, surface roughnessand the design wind speed. The Force Coefficients for such members are to be taken from [Table 27].
9.2.5 Open Plane Frames
Single and mutiple open frames are covered in [6.3.3.3 and 6.3.3.4]. Depending upon the solidity ratio ‘~’,ForceCoefficients data is divided into three categories as:
a) All flat sided members;
b) All circular members having either
i) DVd<6 m2/s
ii) DV~ > 6m2/s
c) Frames with both flat and circular members.
The values of the Force Coefficients are given in [Table 28]. In case of mutiple parallel frames, the windwardframes have a shielding effect on the leeward frames. The wind force on the windward frames or any unshieldedpart of the leeward frames is computed using [Table 28] but the wind load on fully shielded or partly shieldeddepends upon the effective solidity ratio and the frame spacing of the windward frame. [Table 29] gives theShielding Factor ‘q’ in terms of Effective Solidity Ratio ‘~’ and the frame spacing ratio.
9.2.6 Lattice Towers
The data in [6.3.3.5] for overall Force Coefficients for wind blowing into the face of the lattice tower vis-a-visSolidlty Ratio ‘$’ is divided as undec
a) Square towers with flat members,
b) Square towers with rounded members, and
c) Equilateral triangular towers with rounded members.
In case of square towers with flat members, wind is also to be considered blowing into a comer of the tower;for which purpose the Code recommends a multiplying factor of 1.2 for the wind blowing into the face.
9.2.7 Solved Examples by Force Coejicient Method
13xample 16. A hoarding of 2,5 m x 1.5 m at height with its centre 2.5m above ground level is to be erectedon National Highway outside Delhi (Fig. 30).
Given:
Physical Parameters:Height (h) : 1.5mLength (1) : 2.5 mHeight of its centre above the ground: 2.5 m
Wind Data
Wind zone :4 (Basic wind speed= 47 rnh) for DelhiTerrain category :1 [5.3.2.1]Class of structures : A [5.3.2.2], since all dimensions are less than 20 m.
Design Wind Speed (Vz)
VZ = Vb . kl . k2 . k~ [5.3]v~ = 47 lrdsk, = Probability factor for Vb = 47 m [5.3.1 and Table 1]
= 0.71 (considering mean probable design life of structure as 5 years )k2 = Terrain, height and structure size factor [5.3.2 and Table 2] for heights 10 m
= 1.05k~ = Topography factor [5.3.3 and Appendix C]
= 1.0, that is, ground is almost flat.
Hence, VIO= 47 x 0.71 x 1.05x 1.0= 35.04 m/s
91
SP 64 (S & T) :2001
1T.—
1.5K
25 m
L
1.J
I.’c,
30 (a) Pressure Normal to Surface Pressure
m
:;:
t------b
I,.
CfDistribution
30 (b)
nG. 30 HOARDING
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ISP 64 (S & T) :2001
Design Wind Pressure (pZ)[5.4]
pZ = 0.6 x (VZ)2 and PIO= 0.6 x (35.04)2 = 736.68 N/m*Cf = Force coefficient for low wall or hoarding <15 m
Wind Pressure and Forces [6]
Force Coefficients [6.3 and Table 24]
b=2.5mh=l.5mh’= 1.75 m.so h’ > 0.25h and b/h = 1.67
Hence, Cf = 1.2
Effective Frontal Area (Ae) = 2.5 x 1.5= 3.75 m2
Wind Force on Hoarding (F)
F = CP A.. p~
F= 1.2x 3.75x 736.68=3 315.06 Nor 884.02 N/mz
Check for Oblique Winds [6.3.2.3]
The value of Cfin the case is 1.7 on one edge and 0.44 on another edge [Fig. 30 (a)].
For overall forces, Cf is to be considered average of these two values, that is, Cf (average) = 1.284
Wind force (average wind force due to oblique wind)
= 1.284X 3.75X 736.68 N= 3547,11 Nor 945.90 N/m* \
However, the pressure distribution on the hoarding is given in [Fig. 30 (b)] alongwith the eccentricity of the force.
NOTE — The static forceof3547. 11N due to oblique incidence of wind at an eccentricity of 0.25 m from the center of hoarding sheetis to be considered for design of support. The pressure varies on hoarding sheet from maximumof2254.24 N/mz at edges to1085.13 N/mz near central portion, with average pressure of 945.90 N/m*.
Though dynamic wind force may not be experienced by such low structure, the large hoarding may get twisted in strong winds andtearing of sheets may occur near the edges if not designed propwly. Hence, the large hoarding may be braced (in its plane) suitably andmay be stiffened in normal direction (Perpendicular to the plane of hoarding).
Example 17. A masonry freestanding boundary wall (Fig. 31).
Given:
Physical Parameters:
Height of wall above ground level: 3.0 mHeight of fencing (barbed wire) : 0.6 mWidth of wall : 0.345 m uniform heightLength of wall : lkmGround : Almost flatMass density of brick masonry : 17.275kN/m3
Wind Data
Wind zone 2 (Basic wind speed= 39 ink).Terrain category 2Class of structures C [5.3.2.2].
Design Wind Speed (VJ
Vb . kl . kz . k~ [5.3]39 dsProbability factor [5.3.1 and Table 1]0.92 (considering mean probable design life of wall as 25 years)1.05 for terrain category 2 and class of structure C1.0, that is, ground is almost flat.
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J 1200mfni
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BRICKWALL(345mm THICK)
FIG. 31 BOUNDARYWALL(LENGTH=Han)
HenceVIO = 39X 0.92X 0.93X 1.0= 33.37 111/S
Design Wind Pressure (pZ)[5.4]
Pz = 0.6X (VZ)2
PIO = 0.6x(33.37)2= 668.13 N/m2
Wind Load Calculation
Wind force is given byF= Cp A,. p=
Choice of Cf [6.3.2.3 and Table 24]Here, b =1 km, h= 3 mandbfi >160Since wall is on ground Cf = 2.0
Force on Barbed Wire
Solidity ratio = 10 percent, corresponding value of Cf for single frame having flat sided members is 1.9[Table 28]
A.= 0.1 m2
F=l.9x0.1x 668.13 =127N
Force on WallF= 2 x 1 x668.13= 1 336.26 N/m2
Forces at BaseBending moment due to barbed wire and wall
,,
,,
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SP 64 (S & T) :2001
= 127 x 3.525 + 0.5 x 1 336.26x 3.075x 3.225=7 073.44 N-m
Total shear at base= 127+ 1 336.26x 3.075 = 4.236 kN/m length
Calculation of Stresses at Base
Total weight of masonry at base= 17.275x (3.225x1x 0.345)= 19.22 kN/m width of wall
Bending stress caused by vertical bending produced at base on extreme edge= 6M/bd2=6x7073.44/1 x 0.3452= 0.357 N/mm3
Compressive stresses due to weight at base= P/A= 19.22/(1 x 0.345)= 0.055 N/mm2
Maximum tensile stress at extreme fibre for vertical bending= (0.055 - 0.357)= 0.3 N/mm2
Check for Tensile Stress
According to 5.2.3.2 of IS 1905:1980 ‘Code of practice for structural safety of buildings: Masonry walls (secondrevision)’, permissible tensile stress for freestanding vertical walls in vertical bending is 0.07 N/mm2. However,the tensile stress at base of wall at ground level amounts to 0.3 N/mm2, therefore, exceeds the permissible value.The base width of wall is to be increased for tensile stress design. The thickness of wall at various heights isgiven by:
t==h 44008.78 +(7X 104+ 17275h)
Thustatheight 1 m from top= 0.214 mtatheight 2 m from top= 0.392 mtatheight 3 m from top= 0,544 m
NOTES
1 The tensile stress criteria may govern the design. For preliminary design the following formula maybe used to catculate the watlthickness at a height for prismatic walls not exceeding 10 m in height.
r= h J(3 X CfWXfJd) + (7 x 104+ P@)
where
t = thickness wall in m,h = height of section from top of wall (m),CfW=force coefficient for wall,Pd = design wind pressure, andPp = mass density of masonry.
2 The permissible tensile stress in masonry is based on assumption that the mortar used is not leaner than Ml (US = 1/5).
3 Though [Table 1] considers mean probable design life of 5 years for boundary watls but the designer may decide higher k, factor asgiven in [Note of Table 1]dependkrg upon the irnportartcegiven to the structure such as security wall where mean probable design lifemay be taken as 25 years.
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Example 18. A closed cylindrical petrochemical tank (Fig. 32) [6.2.2.8].
ELEVATION
32(a) ~~ for Roof
=-1.2, +0.2
CP=-2.2, -0.8
cp=-1.7, -Q.3
o
~ (Design) for Roof
FIG. 32 A CLOSED CYLINDRICALPETRO-CHEMICALTANKGiven:
Physical Parameters:
Height (h)Diameter (d)Roof angle (et)TopographyExternal surface
:6m: 10m: 11.310: Flat ground: Rough
Wind Data
Wind zoneTerrain categoryClass of structure
4 (Basic wind speed= 44 m/s)2 [5.3.2.1]A [5.3.2.2]
Design Wind Speed (VZ)
Vb = 44 dskl = 1.07 (considering important structure having a return period of 100 years) [5.3.1]k, = 1.0 [5.3.2.2, Table 2 ]k; = 1.0 (site upwind slope c 3°) [5.3.3.1]VIO = 44x 1.07x 1.OX 1.0=47 .08111Ls
Design Wind Pressure (pZ)
p10 = 0.6x (47.08)2= 1329.92 N/m2
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Wind Load Calculation.F = (CPe- CPi)A.pd [6.2.1]
Internal pressure Coefl’icientaon Roof (Cpi) [6.2.3.1 md 6.2.2.8]
As per [6.2.3.1] , the permeability of completed tank may be less than 5 percent and corresponding CPi = t 0.2but during fabrication a stage may come when opening on roof etc maybe equivalent to open ended cylinder.In such a situation for Wd = 0.6 as per [6.2.2.8] CPi=- 0.8. However, there can bean inside positive pressure of+0.7 [6.2.3.2] during construction phase. However, if the tank is pressurized for whatever reason, the workingpressure must be taken for CPi.The design should be checked with C i = -0.8 as long as there is no roof. Once
J“the placing of the roof begins, the internal pressure, should be chang m accordance with [6.2.3.2]. This meanspositive internal pressures of 0.7 can be expected.
External Pressure Coefficients on Roof (CPe) [Fig. 2]
Here, ldd = 0.6 and tan u = 0.2, and [6.2.2.12 and Fig. 2] are applicable.
Therefore, then a = 0.15 h +0.2 D [since 0.2< Wd <2 ] = 2.9 m
Since wind may come from any direction, the CP values are given in [Fig. 32(a)]
Design Pressure Coefficients on Roof (Cp)
Cp= Cpe- Cpi,C’pi= -0.8
The values of CPfor design of roof cladding when the shell is empty are given in [Fig. 32(b)].
Forces on Shell Portion
For component design, pressure coet%cients as described in [Table 18] are to be used.
For overall forces [Table 23] is to be used.
Overall forces on shell (for completed tank)
Here, k’~.b26, hlD = 0.6 and external surface is roughCd = 0.7 [Table 23]A. = 10 (6+1)=70 m2
Force F = Cr A,. p~ = 0.7 x70x 1329.92=65 166.08 N
When tank is open
When tank is open at top, the external pressure distribution around the tank maybe considered as applicable tocylindrical structures described in [Table 18], for ?dD = 1 and different positions of periphery 0, CF valueare:
0 0° 15° 30° 45° 60° 75° 90° 105° 120° 135° 150° 165° 180°Cpe 1.0 0.8 0.1 -0.7 -1.2 –1.6 –1.7 –1.2 -0.7 –0.5 -0.4 -0.4 -0.4
CPi= –0.8 (when the tank is without roof), andCPi= i 0.5 (when the tank is partially open) [6.2.3.2]
The maximum net Cpand force are tabulated below
0 0° 9CF’ 180”
CF +1.0 -1.7 -0.4
Cpi -0.8 -0.8 -0.8
Cp -1.8 -0.9 +0.4
ForceN/m* -2393.86 -2260.86 -1063.94
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Example 19. A circular water tank resting on shajl (Fig. 33).
1182”81 KN .65KNrn?.
Given:
m-t -’27%”N’m
Physical Parameters:
s 1158348N/n
9801.kl N/m
==! 13747.K! N/m
i
12648.09 N/m
11085”01 N/m
I1 9.Om
4 JELEVATION
EO-“A -—.—.~D
PLAN
FIG. 33 WATER TANK RESTINGON SHAIT
Height (h) below top cap : 25.25 mTotal height (H) : 27.00 mDiameter of shaft (outer) : 9.0 mDiameter of tank : 13.0 mSurface condition : Fairly rough on tank and fluted on shaft
Since the height to minimum lateral dimension is less than 5.0, it is not necessary to check the design for dynamicforces [7.1].
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Wind Data
Wind zone : 3 (Basic wind speed= 47 m/s).Terraincategory :1Class of structure :BTopography : Almost flatProbabledesign of structure : 50 years.
Design Wind Pressure at Various Heights (pz= Z)
For calculating design wind pressure at various elevations of the tank only ~ will change and accordingly pZ atith elevations have been computed below,
At 27m elevation
Vb = 47 ndsk, = 1.0kz = 1.03 as per [Note 2 of Table 2]k~ = 1.0VZ = 47X 1.03 =48.41 tn/S
P, = 0.6x (48.41)2= 1406.11 N/m2 between 20 m to 27 m
For 20 m to 15 m height
k2 = 0.98VZ = 47x0.98 x1.OX 1.0=46.06 m/s
P. = 0.6 x (46.06)2= 1272.91 N/m2
For 10 to 15 m height
kz = 0.94VZ = 47x0.94x 1.OX 1.0= 44.18 m/s
P. = 0.6x (44.18)2= 1171.12 N/m2
For Oto 10 m height
kz = 0.88VZ = 47x0.88x l,OX 1.0= 41:36 m/s
P. = 0.6x (41.36)2= 1026.39 N/m2
Wind Load Calculation
Uplift wind pressure on roof [Table 19] with H = Totalheightof the tankportionand D = diameter of the tankportion,
HID= 9/13 = 0.69
CP = –0.78 (by interpolation)and CPi = * 0.2 (for less than 5 percent openings)
Vertical load on roof [6.2.2.9]
P = 0.785 D* \C i -Cd p~ with eccentricity e = O.lDP = 0.785x 13 {6.2 - (-0.78)(1 406.1 1)}
= 182810.75 NEccentricity =0.lD=O.lx13=l.3mBending moment = 1.3x 182.81 kN-m
= 237.65 kN-m
Overall Horizontal Force on Tank and Shaft [6.3.2.1 and 6.3.2.2]
In the case of such appurtenances which ‘Stand’ on another structure, one is confronted with the problem offinding the height to diameter ratio. There is no guideline in the Code as to how such situations are to be handled.A reasonable approach is to take an imagintiy plane at half the height and assume that the airflows symmetrically(or bifurcates symmetrically) about this plane so that this plane can be considered as a virtual ground. Then theheight to be considered is H/2 and diameter is D and the suitable value of height to diameter ratio is (H/2D) , Inthis problem, we have to consider H/2D = 9/(2 x 13)= 0.346.
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SP 64 (S & T) :2001
F = CPAe.p~ [from Table 23 for circular section]V&b 26 and rough surfaceFor tank, H/D = 0.346 which is less than 0.5 and hence, Cf = 0.7
For the shaft, the presence of the wide water tank at the top and the ground at the bottom, may be viewed asmaking the air flow two dimensional that is H/D for the shaft+ CO.Hence it is appropriate to take, Cf = 1.2 forthe rough shaft.
Horizontal Force on Shaft per metre Length
(0.0 to 10.0 m height) = 1.20X(1 X 9.0)X 1026.39= 11085.01 N
(10.0 m to 15 m height) = 1.20x(1 x9.O)X 1171.12= 12648.09 N
(15.0 m to 18 m height) = 1.20x(1x 9.0)x 1272.91
(
(
= 13747.43 N
8.0 m to 19.75 m height) = 0.7x(1X 11.O)X 1272,91= 9801.41 N
9.75 m to 20.00 m height) = 0.70(1X 13)X 1272.91= 11583.48 N
(20.00 m to 25.25 m height) = 0.7X 135X 1406.11= 12795.6 N
(25.25 m to 27.00 m height) = 0,7X (13.0/2)X 1406.11= 6397.80 N
[Appendix D] and its note also specify the value of Cf for circular sections and the procedure is as under. It isto be noted that the pressure at the nearest nodal point of [Table 2] above the region is to be used.
The value of the force coefficient given in [Table 23] for circular cross-section may not appear to be consistentwith the values read from [Fig. 5], and corrected for height to diameter ratio as per the recommendations in[Table 25]. It will be observed that in general the values in [Table 23] are higher even if one assumes that thevalues at finite H/D are those corresponding to the use of multipliers in [Table 25] on the values at H/D + CO.However, from the wording in [Table 23], it is clear that the roughness envisaged in [Table 23] includesdeliberately made small projections, possibly from architectural considerations. This case is not considered in[Fig. 5] The word ‘Smooth’ in [Table 23] really means in this problem E/D= 0,05x 10-5 of [Fig. 5] (for theforce coefficient of 0.6 at H/D + 00). For a diameter of 13.0 m, this means that the combined deviations/rough-ness of surface must be less than 0.65 mm, which is not likely to be achieved in construction.
If the surface is stipulated to be smooth with no fluting or projections, designer may take the values in [Fig. 5]instead of values in [Table 23], taking the multiplier as 0.8 for H/D e 2.0. In the present problem the forcecoefficients then will be 0.64 for tank and 0.8 for shaft.
Example 20. Consider the circular tank of Example 19 to be supported on 16 columns of 45 cm diameter @22.5° radially with a radius of 4.5 m. Other parameters remain unchanged (Fig. 34).
As in the previous problem, the height to maximum width can be taken as 3.0 and hence no dynamic analysis[7.1] is needed.
Force on roof of the dome [6.2.2.9 and Table 19]
Since tank rests on columns, it is covered under Case (d) of [Table 19]
H=9.0m, Z=27.0m, D=13.0mand (2YH-1)=2;The value of CP = -0.75 (for roof) and= -0.6 (for bottom).Since, openings are less than 5 percent, C~i = * 0.2
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WIND -DIRECTION
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945 m 11
ELEVATION
PLAN
FIG. 34 WATER TANK RESTINGON COLUMNS
Uplift Force on Roof per Square Metre
= (Cpe- Cpi)A.Pz= (-0.75 -0.2)x1x 1406.11=+1 335.80 N/m2
Force on Bottom per Square Metre
= (-0.6 -0.2)x 1 x 1 272.91= -1018.33 N/m2
Uplift Force on Roof [Table 19]
P = 0.785 D2 (~pi – CPe.pJ= 0.785X 13 (0.2+ 0.75X 1406.11)— 177214.50 N with an eccentricity ‘e’ = 0.1 D = 1.3 m
Horizontal Force on Tank [6.3.2.1 and 6.3.2.2, Table 231
For HID = 0.347, rough surface, and vd b ~ @Cf = 0.7
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Force on cylinder of tank = 0.7 x(1 x 13) 1406.11 = 12795.60 N/m*
Force on taper= 0.7 (1 x 11.1) 1272.91 = 9890.51 N/m*
Force on staging [6.3, 6.3.3 and 6.3.3.3, Table 28]
Consider the columns as a frame around a periphery and using [6.3.3.3, Table 28]
No. of columns = 16 D.V~ >6 m/s2
Total gap between cylindrical columns (Fig. 34) is
s = [ro sin zz.s”–~ll+{[~.sin45° –~11–[~.sin22.5°+ rll }+ ([r. sin67.5°–rll-[r-()sin 45° + r,] }+( [r. sin 90°-rl] - [r. sin 67.5°+ rl] }
There is no gap for air flow between the columns at El= 90° and El= 67.5°Hence net gapg = rt)sin 67.5° – 6r1 = 4.5 sin 67.5°– 6 x 0.225 = 2.807 m
Solid area = 9.45 – 2.807 = 6.643 m2
Solidity mtio ($)= 6.643/9.45= 0.70
Force coefficient (by interpolation) on wind facing columns= 1.28
Overall Forces on Staging
There is no specific guideline in the Code for finding out the force coefficient for such shielded configuration.A conservative approach is to note that gap flow occurs only for t3 up to 67.5 and hence the ‘frame spacing’as per [6.3.3.4 ] is likely to be closer to unity. The range of values for @r and ~ in [ Fig. 7 ] does not coverthe situation here. Hence, it is conservative to take ‘effective solidity’ as 0.7 and ‘frame spacing ratio’ as 1.0in [ Table 29 ] which yields a shielding factor of 0.6. Hence total force coefficient, Cf = 1.28 + 1.28 x 0.6– 1.28 (1 + 0.6) = 2.048 for combined effect qf windward as well as leeward colums. Here, exposed area, A, is—
6.643 m2 per m length of staging. The wind induced force per unit length of staging FZ = Ct . A. . P~, iscalculated as follows.
FOI- FZ (0.0 m to 10 m height) = 2.048 x 6.643 x 1026.39 = 13963.9 N/mFor F, (10 m to 15 m height) = 2.048x 6.643 x 1171.12 = 15932.93 N/mFor FZ (15 m to 18 m height) = 2.048x 6.643x 1272.91 = 17317.77 N/m
NOTE— Most of the time the columns of the staging are braced with cross-beams. The design of the beam should be able to withstandthe wind loads imposed on it, Special attention is required in detailing of the junction of the beam and the column, For calculation offorces on beams let us consider cross-beams of size 450 mm (depth) and 300 mm (width) which are placed at a vertical sp~cing 013 m. It is more appropriate to adopt a value of Cf from [ Fig. 4A.]. The beam can be considered having parameters as u = 300 mm,b = 450 mm, a / b = 0.67 and n = distance between two columns = r d 8 = 4.5 x 22.5 x 11/180= 1767 mm, It may be noted that theends of the cross-beam are obstructed by columns, therefore, from [ 6.3.3.2 a (ii)] the value of h/b =coisto be adopted. Correspondinglyfrom [Fig, 4A] Cl for windward area is 2.75 and considering the effect of shielded leeward portion the total Cf= 2,75x (1 + 0,6)= 4,4The exposed area of the beam all around the staging at an elevation = area of one beam ( that as half perimeter x width) – area coveredby columns = 0,45 x 11 x 4,5 – 0.45 x 6.643 = 3.372 m2. Therefore, the overall forces acting on a cross-beam spanning all around~he staging at an elevation ‘z’ is given by
For Fz (0.0 m to 10 m height) = 4.4x 3.372x 1026.39= 15228.34 NFor F, (10 m to 15 m height) =4.4x 3.372x 171.12= 17375.67 N
Forces on Each Cross-beam Spanning Two Columns
For calculation of forces on a cross-beam spanning two consecutive columns, the exposed area Ae = 1.0 x 0.45= 0.45 n12/m and Cf = 2.75. Correspondingly forces are as follows:For FZ(0.0 m to 10 m height) = 2.75 x 0.45 x 1 026.39= 1270,16 N/m.For Fz (10 m to 15 m height) =2.75 x 0.45 x 1 171.12= 1449.26 N/m.
Force on Each Column per Unit Length
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Force on each column for calculation of design loads areas follows. Here, Cf = 1.2 [Table 23 ] for an aspectratio ( H/D ) of 18 / 0.45 = 40, V~ . b >6 and for roughr. c. c. column surface, Ae = 1 x 0.45 m2/m length ofcolumn.For Fz (0.0 m to 10 m height ) = 1.2x 0.45 x 1 026.39= 554.25 N/mFor Fz (lOm to 15 m height)= 1.2x 0.45 x 1 171.12 =632.40 N/mFor Fz (15 m to 18 m height)= 1.2x O.45 x 1272.91 =687.37 N/m
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ELEVATION
~@ 2063462 N
~@ 1E770.26 N
~@ 10393.69 N
PLAN
FIG.35 AN INVERTEDTYPE WATER TANK
Example 21. An inverted type conical water tank (Fig. 35) for a nuclear power plant.
Given:
Physical Parameters:
Height (h) :29.0 mTop diameter :14.8 mBottom diameter :3.0 m
Wind Data
Wind zone :3 (Basic wind speed= 44 m/s)Terrain category :2Class of structure : B [5.3.2.2]Topography : FlatProbable life of structure :100 years.
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Design Wind Speed
Vz = VMkl.k2.kJV~ = 44 lllh [5.2]kl = 1.07 [5.3.1]kz = 1.095 [5.3.2.2 and Table 2] (by interpolation)k~ = 1.0 [5.3.3.1]Vz = 44x 1.07X l.095xl.o=51.55 m/s
Design Wind Pressure
For z =29
P.= o.f5 (vz)2Pz = 29= 0.6x (VZ)2 = 0.6 x (51.55)2 = 1594.44 N/m*
The design pressure for other elevations are given in Table 10:
Table 10 Design Pressure at Different Levels
Elevation kl k2 k3 Vz (III/s) pZ (N/m2)
29 1.07 1.095 1.0 51.55 1594.44
24 1.07 1.07 1.0 50.37 1522.28
18 1.07 1.038 1.0 48.87 1432.96
12 1.07 0.99 1.0 46.61 1303.49
6 1.07 0.98 1.0 46.14 1277.34
Force Coeftlcient
Height/Breadth ratio = 29/5 = 5.8; Considering the surface of the tank as rough, V@ 26 and for thisheight/breadthratio the force coefficient Cfas per [Table 23] is 0.8
Wind Load
Wind force F2g = 0.5.Cf . pd . A
= 0.5X 0.8X 1594.44 XA
= 0.5X 0.8X 1 594.44X (14.8+3)X5X 0.5 =28 381.03 N
F2d = 0.8X 1522.28X(3X3+ 22.25)=38 057 N
F18 = 0.8X 1 432.96X(6X3)=20 634.62 N
Flz = 0.8X 1303.49 X(6X3)= 18770.26 N
Fc = 0.8X 1 277.34x(6X3)= 18393.69 N
Reaction, shear force andbending moment assuming the watertankas a cantileverbeam
Reaction at base= 124236.55 N
B.M. and S.F. at an elevation of 24 m = –141 904.9 N-m and28380.98 N respectively
B.M. and S.F. at an elevation of 18 m = –540 532.78 N-m and 66437.98 N
B.M. and S.F. at an elevation of 12m = - 1062968.3 N-m and 87072.6 N
B.M. and S.F. at an elevation of 6 m = –1 698025.5 N-m and 105842.86 N
B.M. andS.F.atbase=–2443 444.8 N-m and 124236.55 N
Example 22. A multistoreyedframed building (Fig. 36).
Given:Physical Parameters:
LengthWidthHeightHeight of each storeySpacing of frames
: 50.0 m: 10m: 60m: 4m: 5 m along the length
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Wind Data
Wind zoneTerrain categoryClass of structureTopographyLife of structure
1 5m 1, Snl L.1 ~ 1
FIG. 36 A MULTISTORY BUILDING
5 (Basic wind speed= 50 rids)3C (since maximum dimension> 50 m)Flat that is upwind slope e 3°50 years.
Design Wind Speed
Vz = V~.kl.kz.kJkx = 1.00
kz = varies with heightk, = 1.00
Vz = (50X 1.00X 1.00) kz = 50.00 kz
SP 64 (S & T) :2001
r
Design Wind Pressure
P. = 0.6 ~ = 1500 k; N/m2
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SP 64 (S & T) :2001
Wind Load Calculations
F=Aexp~xC~
a/b = 10/50= 0.2
h/b = 60/50= 1.2
C~for these values from [Fig. 4] = 1.2
After selecting proper values of kz from the Code, the values of the design wind pressure are computed andgiven in Table 11
Table 11 Wind Load at Different Floor Levels
,Elevation(m) k2 v, (In/s) pZ(N/m2) Lateral
Force(kN)
60 1.04 52.CKI 1622.4 19.5
56 1.03 51.5 1591.5 38.2
52 1.02 51.0 1560.6 37.4
48 1.01 I 50.5 1530.2 36.7 {
44 1,00 50.0 1500.0 36.0
40 0.99 49.5 1470.2 35.3{ ,36 0.975 48.75 1426.0 34.2
32 0.96 48.0 1382.4 33.1
28 0.95 47.5 1353.8 32.6
24 0.93 46.5 1292.4 31.1I I 1
20 0.91 45.5 1242.0 29.8
16 0.87 43.5 1135.0 27.3
12 0.84 42.0 1058.6 25.4
8 0.82 41.0 1008.6 24.2
4 0.82 41.0 1008.6 24.2
0 0.00 0.0 0.0 0.0
Example 23. A 30 m high self supporting square tower (Fig. 37).
Given:
Physical Parameters:
Height of tower : 30.0 mBase width : 5.0 mTop width : 1.50mNumber of panels :8Panel height : Top five 3.0 m and rest three are 5.0 m eachType of bracings : K-type with apex upwards in 5 m panels and X-type in 3 m panelMember properties : Table 12
Wind Data
Wind zone :4 (Basic wind speed= 47 mh)Life of structure :100 yearsTopography : Flat
-1
,,
/
Design Wind Speed
Vz = V~.kl .kz.kJk~ = 1.36kz = varies with height
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Table 12 Member Properties
Elevation(m) Member Size(mm) No. & Length (m)
30-27 Horizontal 65x65x6 1, 1.5
Vertical 90x90x6 1,3.0
Diagonal 100x 1OOX6 1,3.54
Plan Bracing-3 65x65x6 1,2.121
27-24 Horizontal 65x65x6 1, 1.5
Vertical 90x90x6 1,3.0
Diagonal 100x 1OOX6 1,3.54
Plan Bracing-3 65x65x6 1,2.121
24-21 Horizontal 65x65x6 1, 1,5
Verrical 65x65x6 2,3.008
Diagonal 100x IOOX6 1,3.464
Plan Bracing-3 65x65x6 1,2.121
21-18 Horizontal 65x65x6 I, 1.937
Vertical 65x65x6 2,3.008
Diagonal 65x65x6 2,3.701
Plan Bracing-3 80X80X6 1,2.740
18-15 Horizontal 65x65x6 2,2.375
Vertical 75x75x6 2,3.008
Diagonal 65x65x6 2,3.972
Plan Bracing-3 IOOX 100x6 1,3.359
15-10 Horizontal 65x65x6 2,(1.406x2)
Verrical 75x75x6 2,5.027
Diagonal 90x90x6 1,5.317
Plan Bracing-2 65x65x6 1,1.989
Plan Bracing-1 65x65x6 1,2.812
10-5 Horizontal 65x65x6 2,(1.771x2)
Vertical 90x90x6 2,5.027
Diagonal 90x90x6 1,5.449
Plan Bracing-2 75x75x6 1,2.504
Plan Bracing-1 65x65x6 1,3.542
5-GL Horizontal 65x65x6 2. (2.135x 2)
Vertical 1IY3x1OOX6 2,5.027
Diagonal 100x 1CN3X6 1,5.602
Plan Bracing-2 90x90x6 1,3.02
Plan Bracing-1 65x65x6 1,4.271
k, = 1.07Vz = (47X 1.07X 1.36.) k2 = 68.39 kz
Design Wind Pressure
PZ = 0.6(VZ)2=2806.32 k: N/m2
After selecting proper values of k2 from the Code, the values of the design wind pressure are computed andgiven in Table ~3. -
Computation of Solidity Ratio (0)
Enclosed area of the top half panel that is panel between 15 m and 13.5 m height
A = 1.50x 3.0/2= 2.254 m2
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Solid area (Ao)= Area of legs + Area of horizontal members +Area of braces= 0.7029 m2
Add 10 percent for gusset platesetc=0.7029 + 0.0703 = 0.7732 m2
Solidity ratio($) = 0.773 2/2.25= 0.344
Force coefficient for square lattice tower with flat members [6.33.5 and Table 30]
Cf between elevation 30 m and 28.5 m elevations = 2,58 (by interpolation)
Wind Load Calculations
F= Aexp~x C~
Fat elevation 30 m = 0.7732 x 2 806.32x 1.13x 1.13,x 2.58=7 150.06 N
Shear Force Calculations
Shear force at elevation 30 m = 7150.06 N
Bending Moment Calculations
Bending moment at elevation 30 m = 0.00
Bending moment at elevation 27 m = 7 150.06x 3.0=21 450.18 N-m
The panel loads, shear forces and bending moments have been computed for different elevations are given inTable 13.
Table 13 Computation of Velocity, Pressure, Drag Coefflcien$ Shear Force and Bending Moments —Wind Normal to any Face
kl = 1.07, k~ = 136, Vb = 47m/s
No. Elevation k2 VA (m/S) PA (Nlm2) Ae (mz) 4 Cf Panel SF (N) BM (N-m) 1Load(N)
1. 30 1.13 77.28 3584.25 0.7732 0.344 2.392 7150.06 7150.0 0.0
2. 27 1.21 76.67 3526.97 1.4390 0.320 2.462 13703,34 20853.4 21450.18
3. 24 1.112 76.05 3470.16 1.5860 0.3285 2.437 14623.26 35476.7 84010.4
4. 21 1.103 75.44 3414.72 1.9100 0.3287 2.436 17348.83 52825.5 190440.4
5. 18 1.088 74.44 3322.11 2.1935 0.3080 2.497 20112.25 72937.8 348916.8
6. 15 1.07 73,18 3213.19 3,1476 0.2727 2.624 29694.2 102632.0 567730.1
7. 10 1.03 70.45 2977.92 3.1830 0.1797 3.ot13 32237’.13 134869.1 1080890.0
8. 5 1.03 70.45 2977.92 5.4020 0.1617 3.080 56158.75 191027.9 1755235.0
9. 0 1.03 70.45 2977.92 — — — — 191027,9 2710375.0
9.3 Gust Factor Method
9.3.1 Methodology of computationof wind load by Gust Factor Method has been explained in 6. Therefore,solved examples of structures have been given below to demonstrate the usage of the method:
9.3.2 Solved Examples Using Gust Factor Method
Example 24. Tofind design wind pressure for a rectangular clad building.
Given:
Physical Parameters:
. .
,,
Building : 52 m long, 8 m wide and 40 m highLife of structure : 50 yearsTerrain category :3Topography : FlatLocation : Calcutta
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0“
FIG. 37 LATTICE MICROWAVE TOWER
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Previous Code [1S 875:1964] [Fig 1A]
pZ=210 kgf/m2 = 1960.7 N/m2 at 30 m height,
Present Code [1S 875 (Part 3) : 1987]
As per [7.1] of this Code, since the ratio of height to least lateral dimension (40/8) is 5, dynamic analysis isrequired.
v~ = 50 mlsk, = 1.0 [Table 1, N = 50]k2’ = 0.5 [Table 33]kx = 1.0
Design Hourly Mean Wind Speed
V,’ = 50(1.0)(0.5)(1.0)= 25.0 m/s
Design Wind Pressure
P.’ = 0.6 (25.0)2= 375.0 N/m2
Computation of Gust Factor (G) [8.3]
G = 1 +gpr~[B(l +@2+SE/’~]
For the present case, from [Fig. 8]
gf. r= 1.7, L(h)= 1 150m, CY= 10; CZ= 12
Hence, ~= CYb/CZ h = 10(8.0)/12(40.0)= 0.1667
AlsoC, h /L(h) = 12(40.0)/1 150= 0.4174
Background turbulence factor B = 0.76 [Fig. 9]
Calculation of reduced frequency
To calculate this parameter, one needs to know the natural frequency (/) in the ‘first mode’. Using the expression(b) of [Note 1 of 7.1]
T= 0.09H~= 0.09 X 40.0/~= 1.73 S
f= 0.578 Hz
F. [Fig. 10]= Cz~Oh / V(h) = 12(0.578) (40,0)/33.5= 11117
Size reduction factor (S) from [Fig. 10] is 0.15
The parameter,f~ L(h) W(h)= 0.78 x 1 150/33.5= 26.78
The gust energy factor E from [Fig. 11] is then 0.07
Since the height of the building exceeds 25 m in terrain category 3, @ = O [8.3]. With the damping coefficient~= 0.016 [Table 34],
G = 1 + 1.7 [0.76(1 + 0)2+ 0.15(0.07)/0.016}= 1 + 2.519= 3.519
Hence the design wind pressure = 3.023 (375.0)= 1319.7 N/m2 at 40 m height.
Proceeding similarly, the design wind pressure for larger life and other terrains were considered and the resultsare given below at 30 m height
,.
Location Life Years Terrain IS 875 IS 87.5:(Part-3) : 1964
1987
Calcutta 50 3 1689.2
100 3 1970.28 1960.00
100 2 3002.08
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The large increase in the design pressure for Terrain Category 2 which is very common for tall industrialstructures outside cities is to be noted.
Example 25. A mutistoreyedfiamed building of Exzzmple 22.
Given:
Physical Parameters:
Building : 50 m long, 10 m wide and 60 m highLife of structure : 50 yearsTerrain category :3Topography : FlatLocation : Bhubneshwar \
Design Wind Velocity
As per [7.1] of this Code, Ratio of height to least lateral dimension (60/10) is 6> 5; dynamic analysis is required.
T= 0.09Hld = 0.09 X 60/~= 1.7sFrequency (j) = 1/1.7= 0.58 Hz < 1.0 Hz; dynamic analysis is required
v~ = 5om/sk, = 1.0 [Table 1, N = 50]kz’ = 0.5 [Table 33]k~ = 1.0
Design Hourly Mean Wind Speed
V; = 50 (1.0)(0.5)(1.0)= 25.0 111/S
Design Wind Pressure
p,’ = 0.6 (25.0)2= 375.0 N/m2
Computation of Gust Factor (G) [8.31
G=l+gPr~[B(l +@)2+SE@l
For the present case, from [Fig. 8]
For 60 m height, gP r= 1.5, L(h)= 1200 m, CY= 10; C,= 12
Hence, ~ = CYb/CZ h = 10(10.0)/12(60.0) = 0.138
AlsoCZh/L(h)= 12(60.0)/1 200= 0.60
Background Turbulence Factor B = 0.77 [Fig. 9]
Calculation of Reduced Frequency
FO[Fig. 10]= CZfJ / V(h) = 12(0.58) (60.0)/25= 16.7
Size reduction factor (S) from [Fig. 10] is 0.28
The parameter f$ L(h)/V(h)= 0.58 x 1200/25= 27.84
The gust energy factor E from [Fig. 11] is then 0.07
Computationof Gust Factor
$ = O [8.3]. With the damping coefficient@= 0.016 [Table 34],
G = 1 + 1.5 ~[0.77 (1 + 0)2+ 0.28 (0.07)/0.016] = 3.11
Computation of wind load at different elevations of the building are given in Table 14.
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Table 14 Wind Load at Different Elevations
Level Elevation(m) kg’ v; (In/s) PZ’ Force(kN)
1 60 0.72 36.0 777.6 28,97
2 56 0.71 35.5 756.2 56.40
3 52 0.70 35.0 735.0 54.25
4 48 0,688 34.4 710.0 52.40
5 44 0.676 33.8 685.5 50.80
6 40 0.664 33.2 661.4 48.80
7 36 0.652 32.6 637.6 47.00
8 32 ‘ 0.640 32.0 614.4 45.30
9 28 0.622 31.1 580.3 42.80
10 24 0.606 30.3 540.0 39.80
11 20 0.590 29.5 522.0 38.50
12 16 0.550 27.5 453.8 33.70
13 12 0.520 26.0 405.6 30.30
14 8 0.500 25.0 375.0 28.00
15 4 0.500 25.0 375.0 28.00
16 0 O.ooa 00.0 Ooo.o CQ.00
.
Example 26. The inverted type water tank of Example 21 using Gust Factor Method.
Given:
Fundamental frequency of the structure: 0.49 Hz
For the 29 m height
Vz’ = V~.kl .k2.k~
kl = 1.07k~ = 1.00kz = 0.786VZ’ = 44X 1.07X 0.786X 1.00= 37.00 Illh
Wind Load
F= = @te.p5’.G
Pz = 0.6(37) = 821.4 N/m*A, = 0.5(14.8+ 3)5x 0.5= 22.25 m2Cf = 0.8
Gust factor G = I+gf. r~[ll(l + 0)2 + SE@lg~.r= 1.29 [Fig. 8](3 = 0.016c, = 10, b=6, C,=12, h=29
CY.b/CZ.h=#X$=0.1724
CZ.h/L(h) = ~ = 0.309
For these values, B = 0.8 from [Fig. 9]
Size reduction factor (FO)= CZ.fO. h I V{ = 12x 0.49x 29/37= 4.61
For these values, from [Fig. 10] S = 0.45
Calculation of E (Gust energy factor)= 0.49x 1 125/37= 14.89
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E from [Fig. 11] = 0.09@ = Oin this caseG = 1+1.29 ~[0.8 + (0.45 X0.09/0.016)] = 3.35
Force = 0.8x 22.25 x 821.4x 3.35 =48 992.00 N
For the 24 m height
Vb = 44 mlsk, = 1.07kl = 0.766k~ = 1.00Vz’ = 44x 1.07 x 0.766 x 1.00= 36.06 m/sp,’ = 0.6 (36.06)2= 780.19 N/m*
Force = 780.19x 0.8x 3.35x (22.25+ 9)=65 340.91 N
For the 18 m height
kz = 0.738Vz’ = 44X 1.07X 0.738 X 1.00= 34.74 111/S
,= 0.6 (34.74~ = 724,12 N/m2
F%e = 724.12 x0.8 x 3.35x(9+ 9)=34 931.55 N
For the 12 m height
kz = 0.69V,’ = 44x 1,07x 0.69x 1.00= 32.48 m/s
f = 0.6 (32.48)Z= 632.97 N/m*F%e = 632.97 x 0.8x 3.32x(9+9)=30 261.03 N
For the 6 m height
k~ = 0.67V,’ = 44x 1.07x0.67x 1.00 =31.54 m/s
,= 0.6 (31,54)2= 596.86 N/m2
F%e = 596.86 x 0.8x 3.35x 18=28 792.526 N
Treating the water tank as a cantilever beam, support reaction
R~ = 208318.02 N
B.M. and S.F. at a section x metres from point B
Reaction b’, =48 992.00 NMoment A4x = –FB xIf.r=o A4x=o
lfx= 5 MX=–48 992x 5 =–244960 N-m
B.M. at a section x metres from point F
v, =——
M, =Ii- x =
M, =——
FB + FF48 992+65 340.91 = 114332.91 N-[FB (5 + .x) + FF(x)]6m-[48 992 X11+ 65 340.91X 6]-[538 912 + 392 045/46]= -930957.46 N-m
-1),,
:;_ . . . .
Ifr;4’:.--,%$, ,
,,
.,‘ .,,
,!
,,,
<
B.M. and S.F. at a section x metres from point E
V,= 48 992+65 340.91 +34 931.55= 149264.46 NMx = -[FB(ll+X) +FF(6+X)+ FE(X)]
Ifx = 6mM. = -[48 992X 17+65 340.91 X 12+ 34931.55 x6]
= -[832 864+ 784 090.92+ 209 589.3]= -1826544.2 N-m
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B.M. and S.F. at a section x metres from point D
VX= FD + FE + FF + F’B= 179525.49 NM, = -[48 992 (17 -+x)+65 340.91 (12 +x)+34 931.55(6 +x)+30 261.03x]
If x = 6mM. = -[1 126 816+ 1176136.3 +419 178.6+ 181 566.48] =-2 903697.3 N-m
B.M. and S.F. at a section x metres from point C
Vx = 208318.02 NMX = -[48 992 (23+x)+65 340.91 (18+X)+34 931.55 (12+X)+30 261.03 (6+ x)+28 792.53x]
If x = 6mM, = -[1 420768 + 1568 181.8+ 628 767.9+ 363 132.36+ 172 755.18]
= –4 153605.2 N-m
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